In a ray diagram of a concave mirror, which point is the focal point?

Correct Option: B. Solution: In a concave mirror, the focal point is labeled as 'F'.

Ray Optics and Optical Instruments

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Understand the basic principles of ray optics and optical instruments.
Analyze the behavior of light through reflection and refraction.
Calculate the focal lengths of various lens systems.
Determine the magnifying power of optical instruments such as microscopes and telescopes.
Apply the mirror equation to deduce properties of images formed by concave and convex mirrors.
Explore the concept of total internal reflection and its applications in optical fibers.
Investigate the effects of lens combinations on image formation and magnification.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Nine: Ray Optics and Optical Instruments

9.1 Introduction

Nature has endowed the human eye (retina) with the sensitivity to detect electromagnetic waves within a small range of the electromagnetic spectrum.
Electromagnetic radiation in this region (wavelength of about 400 nm to 750 nm) is called light.
Light travels with enormous speed and in a straight line.
The speed of light in vacuum is approximately c = 3 x 10^8 m/s.

9.2 Key Concepts

Focal Length: The distance from the lens to the focal point.
Magnification: The ratio of the size of the image to the size of the object.

9.3 Exercises

9.1

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm.
Determine the distance from the mirror for a sharp image.

9.2

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm.
Find the location of the image and the magnification.

9.3

A tank filled with water to a height of 12.5 cm shows an apparent depth of a needle at 9.4 cm.
Calculate the refractive index of water.

9.4

Refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface.

9.5

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.
Determine the area of the surface of water through which light from the bulb can emerge.

9.4 Important Formulas

Magnifying Power of a Simple Microscope:
- m = 1 + (D/f)
- Where D = 25 cm (least distance of distinct vision), f = focal length of the convex lens.
Magnifying Power of a Compound Microscope:
- m = me × To
- Where me = 1 + (D/f) (magnification due to eyepiece) and To is the magnification produced by the objective.
Magnifying Power of a Telescope:
- m = B = fo/fe
- Where fo and fe are the focal lengths of the objective and eyepiece, respectively.

9.5 Common Mistakes and Exam Tips

Ensure to differentiate between magnification in absolute size and angular magnification.
Remember that the effective focal length of a lens system can change based on the arrangement of the lenses.
Be cautious with the signs of focal lengths when using the lens formula.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Ray Optics and Optical Instruments

Common Pitfalls

Misunderstanding Image Formation: Students often think that an image does not exist if a screen is removed. Remember, rays converge to form an image even without a screen.
Confusing Magnification and Angular Magnification: Be clear about the difference between absolute magnification and angular magnification (magnifying power). They are not always the same.
Ignoring Sign Conventions: When applying formulas, ensure you are following the correct sign conventions for distances and heights.

Tips for Success

Practice Ray Diagrams: Drawing ray diagrams can help visualize how light behaves with different optical instruments and improve understanding of image formation.
Understand the Formulas: Familiarize yourself with the key formulas for magnification, focal lengths, and the laws of reflection and refraction. Knowing when to use each formula is crucial.
Work Through Exercises: Engage with exercises that differentiate between types of lenses and mirrors, as well as their respective image characteristics. This will solidify your understanding.
Review Critical Angles: Pay attention to the concept of critical angles and total internal reflection, as these are common areas of confusion.
Clarify the Use of Optical Instruments: Understand how different optical instruments like microscopes and telescopes work, including their configurations and magnifying powers.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

Correct Answer: A

Solution:

The magnification

M

produced by a lens is given by

M = 1 + \frac{D}{f}

, where

D

is the least distance of distinct vision (25 cm) and

f

is the focal length (9 cm). So,

M = 1 + \frac{25}{9} = 1

when the object is at the focal point.

1.5 m

2 m

3 m

1 m

Correct Answer: A

Solution:

To project an image on a wall 3m away, the maximum possible focal length of the lens is 1.5 m.

Real and inverted

Virtual and upright

Real and upright

Virtual and inverted

Correct Answer: A

Solution:

For a concave mirror, when an object is placed between the focal point (F) and the center of curvature (C), a real and inverted image is formed beyond C. Here, the object is placed at 15 cm, which is between F (10 cm) and C (20 cm).

5 cm

7 cm

10 cm

12 cm

Correct Answer: A

Solution:

The apparent shift

\Delta h

is given by

\Delta h = t \left(1 - \frac{1}{n}\right)

, where

t = 15

cm and

n = 1.5

. Thus,

\Delta h = 15 \left(1 - \frac{1}{1.5}\right) = 15 \times 0.3333 = 5

cm.

15 cm in front of the mirror

20 cm in front of the mirror

30 cm behind the mirror

40 cm behind the mirror

Correct Answer: B

Solution:

Using the mirror equation

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

, where

f = -10

cm (since it's a concave mirror),

u = -30

cm (object distance is negative in mirror conventions), we solve for

v

\frac{1}{-10} = \frac{1}{v} + \frac{1}{-30}

. Solving gives

v = -20

cm, meaning the image is 20 cm in front of the mirror.

Correct Answer: B

Solution:

The magnification

M

is given by

M = 1 + \frac{D}{f}

, where

D = 25

cm is the least distance of distinct vision. Thus,

M = 1 + \frac{25}{9} \approx 2.78

, rounded to 2x.

1.5 meters

3 meters

4.5 meters

6 meters

Correct Answer: A

Solution:

The maximum possible focal length of the lens is half the distance between the object and the image, which is 1.5 meters.

They bend towards the normal

They bend away from the normal

They do not bend

They reflect back into the denser medium

Correct Answer: B

Solution:

When light travels from a denser medium to a rarer medium, it bends away from the normal.

The object appears closer.

The object appears further away.

The object appears at the same position.

The object disappears.

Correct Answer: A

Solution:

A glass slab causes the object to appear closer due to the refraction of light.

Real and inverted

Virtual and upright

Real and upright

Virtual and inverted

Correct Answer: B

Solution:

When an object is placed between the focal point (

F

) and the pole (

P

) of a concave mirror, the rays after reflection appear to diverge from a point behind the mirror, forming a virtual and upright image.

41.8°

48.6°

36.9°

45.0°

Correct Answer: A

Solution:

The critical angle

\theta_c

is given by

\sin \theta_c = \frac{1}{n}

, where

n

is the refractive index of the denser medium (glass). Thus,

\sin \theta_c = \frac{1}{1.5} \approx 0.6667

. Therefore,

\theta_c = \sin^{-1}(0.6667) \approx 41.8°

19.5°

20°

22°

25°

Correct Answer: A

Solution:

Using Snell's law,

n_1 \sin i = n_2 \sin r

, where

n_1 = 1

(for air),

i = 30°

, and

n_2 = 1.5

. Solving for

r

, we get

\sin r = \frac{1}{1.5} \sin 30° = \frac{1}{1.5} \times 0.5 = \frac{1}{3}

. Thus,

r = 19.5°

It becomes smaller.

It remains the same.

It becomes larger.

It becomes zero.

Correct Answer: C

Solution:

When light travels from a denser to a rarer medium, it bends away from the normal, making the angle of refraction larger.

41.8°

42.8°

43.8°

44.8°

Correct Answer: A

Solution:

The critical angle

\theta_c

is given by

\sin \theta_c = \frac{n_2}{n_1}

, where

n_1 = 1.5

(glass) and

n_2 = 1.0

(air). Thus,

\sin \theta_c = \frac{1}{1.5} = 0.6667

. Therefore,

\theta_c = \arcsin(0.6667) \approx 41.8°

It always produces a real image.

It always produces a virtual image.

It can produce both real and virtual images.

It produces an image larger than the object.

Correct Answer: B

Solution:

A convex mirror always produces a virtual image, regardless of the object's position.

250

100

150

Correct Answer: A

Solution:

The magnifying power

M

of a compound microscope is given by:

M = \left(\frac{v_o}{u_o}\right) \times \left(1 + \frac{D}{f_e}\right)

where

v_o

is the image distance from the objective,

u_o = 9.0

mm,

f_e = 2.5

cm, and

D = 25

cm (least distance of distinct vision). Assuming

v_o \approx f_o = 8.0

mm for maximum magnification:

M = \left(\frac{8.0}{9.0}\right) \times \left(1 + \frac{25}{2.5}\right) = \left(\frac{8}{9}\right) \times 11 = \frac{88}{9} \approx 250

C7H16

C6H14

C8H18

C5H12

Correct Answer: A

Solution:

Isoheptane is a branched alkane with the chemical formula C7H16.

1.5 m

2.0 m

3.0 m

4.0 m

Correct Answer: A

Solution:

The maximum possible focal length of the lens required is 1.5 m.

10 cm

15 cm

20 cm

25 cm

Correct Answer: B

Solution:

Using the lens formula and given distances, the focal length is calculated to be 15 cm.

30°

45°

60°

75°

Correct Answer: B

Solution:

Using the critical angle formula and given refractive index, the angle of incidence is calculated to be 45°.

20x

24x

30x

36x

Correct Answer: B

Solution:

The magnifying power of a telescope is given by the ratio of the focal length of the objective to the focal length of the eyepiece, which is 24x.

\sin^{-1}(\frac{1}{1.5} \sin i_1)

\cos^{-1}(\frac{1}{1.5} \sin i_1)

\tan^{-1}(\frac{1}{1.5} \sin i_1)

\cot^{-1}(\frac{1}{1.5} \sin i_1)

Correct Answer: A

Solution:

Using Snell's Law,

n_1 \sin i_1 = n_2 \sin r_1

, where

n_1 = 1

(air) and

n_2 = 1.5

(glass). Solving for

r_1

gives

\sin r_1 = \frac{1}{1.5} \sin i_1

, so

r_1 = \sin^{-1}(\frac{1}{1.5} \sin i_1)

1.5 m

3.0 m

0.75 m

2.0 m

Correct Answer: A

Solution:

To form an image on the opposite wall 3m away, the focal length

f

of the lens must be such that the image distance

v = 3

m. Using the lens formula

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

, and assuming the object is at infinity, we have

\frac{1}{f} = \frac{1}{3}

, giving

f = 1.5

1500

150

0.15

Correct Answer: A

Solution:

Angular magnification (

M

) of a telescope is given by the formula

M = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}

. Here,

f_{\text{objective}} = 15

m = 1500 cm and

f_{\text{eyepiece}} = 1

cm, so

M = \frac{1500}{1} = 1500

Real and inverted

Virtual and upright

Real and upright

Virtual and inverted

Correct Answer: B

Solution:

For a concave mirror, when the object is placed between the focal point and the mirror, the image formed is virtual and upright.

1500

150

1.5

Correct Answer: A

Solution:

Angular magnification is given by the ratio of the focal lengths of the objective and eyepiece. Thus, it is 1500.

41.8°

45°

48.6°

50°

Correct Answer: C

Solution:

For total internal reflection to occur, the angle of incidence

i

must be such that the angle of refraction

r

at the other face is 90°. Using Snell's law,

n \sin i = \sin r

, where

n = 1.524

and

r = 90°

. Thus,

\sin i = \frac{1}{1.524}

, giving

i = 48.6°

It bends towards the normal.

It bends away from the normal.

It continues in a straight line.

It reflects back into the original medium.

Correct Answer: A

Solution:

When a light ray enters a denser medium, it bends towards the normal due to refraction.

2.5 cm

3.3 cm

4.0 cm

1.5 cm

Correct Answer: B

Solution:

The lateral shift

d

is given by

d = t \cdot \frac{\sin(i_1 - r_1)}{\cos r_1}

. Assuming small angles,

\sin(i_1 - r_1) \approx i_1 - r_1

and

\cos r_1 \approx 1

. Thus,

d \approx t \cdot (i_1 - r_1)

. Given

t = 5

cm and

n = 1.5

r_1 = \frac{i_1}{n}

. For small

i_1

d \approx 5 \cdot (i_1 - \frac{i_1}{1.5}) = 5 \cdot \frac{0.5i_1}{1.5} \approx 3.3

cm.

2.5 cm

3.0 cm

4.0 cm

5.0 cm

Correct Answer: A

Solution:

To find the object distance for the objective lens, we use the lens formula:

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

. Here,

f = 2.0

cm,

v = -25

cm (since the image is virtual and on the same side as the object), and

u

is the object distance we need to find. Solving,

\frac{1}{2.0} = \frac{1}{-25} - \frac{1}{u}

. Solving for

u

, we get

u \approx 2.5

cm.

50.3°

60.0°

40.5°

45.0°

Correct Answer: C

Solution:

The critical angle

\theta_c

can be found using

\sin \theta_c = \frac{n_2}{n_1}

, where

n_1 = 1.68

and

n_2 = 1.44

. Thus,

\sin \theta_c = \frac{1.44}{1.68} = 0.8571

. Therefore,

\theta_c = \arcsin(0.8571) \approx 40.5°

60 cm (converging)

-60 cm (diverging)

12 cm (converging)

-12 cm (diverging)

Correct Answer: D

Solution:

The effective focal length (

F

) of two lenses in contact is given by

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}

where

f_1 = 30

cm and

f_2 = -20

cm. Calculating gives

\frac{1}{F} = \frac{1}{30} - \frac{1}{20} = \frac{1}{60}

, so

F = -60

cm, indicating a diverging system.

Correct Answer: B

Solution:

The magnifying power

M

of a compound microscope when the final image is at infinity is given by

M = \frac{L}{f_o} \cdot \frac{D}{f_e}

, where

L

is the separation between the lenses,

f_o

is the focal length of the objective,

f_e

is the focal length of the eyepiece, and

D

is the least distance of distinct vision (25 cm). Thus,

M = \frac{15}{2} \cdot \frac{25}{6.25} = 7.5 \cdot 4 = 30

1.5

1.33

1.25

1.10

Correct Answer: A

Solution:

Using Snell's law, the refractive index is calculated as the ratio of the sine of the angle of incidence to the sine of the angle of refraction, which is approximately 1.5.

20 cm

30 cm

40 cm

50 cm

Correct Answer: A

Solution:

Using the lens formula for lenses in contact:

\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}

. Solving for

f_2

gives 20 cm.

10 cm

15 cm

20 cm

25 cm

Correct Answer: A

Solution:

The combined focal length of the lens system is calculated using the lens formula for lenses in contact, resulting in 10 cm.

2.0 cm

2.5 cm

3.0 cm

4.0 cm

Correct Answer: A

Solution:

For the final image to be at infinity, the object should be placed at the focal length of the objective lens. Thus, the object should be placed 2.0 cm from the objective lens.

It is the point where parallel rays converge after reflection.

It is the point where rays diverge after reflection.

It is the point where the object is placed.

It is the point where the image is always virtual.

Correct Answer: A

Solution:

In a concave mirror, parallel rays converge at the focal point after reflection.

60.0°

50.5°

45.0°

30.5°

Correct Answer: B

Solution:

The critical angle

\theta_c

can be found using the formula

\sin \theta_c = \frac{n_2}{n_1}

, where

n_1 = 1.68

and

n_2 = 1.44

. Thus,

\sin \theta_c = \frac{1.44}{1.68} \approx 0.8571

. Therefore,

\theta_c \approx 50.5°

60 cm

40 cm

45 cm

50 cm

Correct Answer: B

Solution:

Using the lens formula

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

, where

f = 20 \text{ cm}

and

u = -30 \text{ cm}

. Solving for

v

, we get

\frac{1}{20} = \frac{1}{v} + \frac{1}{30}

. This simplifies to

\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}

, hence

v = 60 \text{ cm}

. Therefore, the image is formed 60 cm from the lens.

Correct Answer: A

Solution:

The magnifying power

M

of a telescope is given by

M = \frac{f_o}{f_e}

, where

f_o

is the focal length of the objective lens and

f_e

is the focal length of the eyepiece. Thus,

M = \frac{144}{6} = 24

41.81°

42.00°

43.00°

44.00°

Correct Answer: A

Solution:

For total internal reflection, the angle of incidence should be greater than the critical angle. The critical angle

\theta_c

is given by

\sin \theta_c = \frac{1}{n} = \frac{1}{1.524} \approx 0.656

, so

\theta_c = \arcsin(0.656) \approx 41.81°

2.5 cm

3.0 cm

4.0 cm

5.0 cm

Correct Answer: B

Solution:

Using the lens formula and the given distances, the object should be placed 3.0 cm from the objective lens.

15 cm

20 cm

25 cm

30 cm

Correct Answer: B

Solution:

Using the lens formula

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

and the fact that the image distance

v

and object distance

u

add up to 90 cm, we have two equations:

\frac{1}{15} = \frac{1}{v} - \frac{1}{u}

and

v + u = 90

. Solving these gives two solutions for

u

and

v

, leading to a separation of 20 cm between the two positions of the lens.

19.47°

20.00°

22.02°

25.00°

Correct Answer: A

Solution:

Using Snell's law:

n_1 \sin i = n_2 \sin r

. Solving for

r

gives

\sin r = \frac{\sin 30°}{1.5}

, which results in

r \approx 19.47°

Correct Answer: B

Solution:

In a concave mirror, the focal point is labeled as 'F'.

15 cm

20 cm

25 cm

30 cm

Correct Answer: A

Solution:

Using the lens formula and the given conditions, we can set up the equations for the two positions of the lens. The separation between the two positions of the lens is 20 cm, and the total distance between object and image is 90 cm. Solving the equations, we find the focal length

f = 15

cm.

At the focal point

Beyond the center of curvature

Between the focal point and the mirror

At infinity

Correct Answer: C

Solution:

An object placed between the focal point and the mirror produces a virtual and enlarged image between the focal point and the mirror.

They diverge

They converge at the focal point

They remain parallel

They reflect back

Correct Answer: B

Solution:

A convex lens converges parallel rays of light to its focal point.

28.1°

30.0°

32.1°

35.0°

Correct Answer: A

Solution:

Using Snell's law,

n_1 \sin i = n_2 \sin r

, where

n_1 = 1

(air) and

n_2 = 1.5

. Solving for

r

, we find

\sin r = \frac{\sin 45°}{1.5}

, which gives

r \approx 28.1°

41.8°

42.2°

43.6°

44.0°

Correct Answer: A

Solution:

The critical angle

\theta_c

is given by

\sin \theta_c = \frac{1}{n}

, where

n = 1.524

. Thus,

\theta_c = \sin^{-1}(\frac{1}{1.524}) \approx 41.8°

Greater than 1

Less than 1

Cannot be determined

Correct Answer: A

Solution:

When the angle of incidence is equal to the angle of emergence, the refractive index is 1, indicating no refraction.

12 cm, converging

12 cm, diverging

-12 cm, converging

-12 cm, diverging

Correct Answer: D

Solution:

The effective focal length

f

of two lenses in contact is given by

\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}

. Here,

f_1 = 30

cm and

f_2 = -20

cm. Thus,

\frac{1}{f} = \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} = -\frac{1}{60}

. Therefore,

f = -60

cm, indicating a diverging system.

Correct Answer: B

Solution:

The magnification produced by the lens is 2x since the lens is used at its focal length.

7.5 cm

8.5 cm

9.5 cm

10.5 cm

Correct Answer: A

Solution:

To find the distance between the object and the lens, we use the magnification formula and the given area of the virtual image. Solving for the object distance, we find it to be approximately 7.5 cm.

8 cm

9 cm

10 cm

11 cm

Correct Answer: B

Solution:

The magnification

M

is given by

M = \frac{A_{image}}{A_{object}} = \frac{4}{1} = 4

. For a magnifying glass,

M = 1 + \frac{D}{f}

, where

D

is the distance of distinct vision (25 cm). Solving

4 = 1 + \frac{25}{10 - v}

, we find

v = 9 \text{ cm}

True or False

Correct Answer: True

Solution:

Optical density is related to the refractive index, which is the ratio of the speed of light in two media.

Correct Answer: False

Solution:

Optical density should not be confused with mass density. The refractive index is related to the speed of light in the medium, not its optical density.

Correct Answer: False

Solution:

The structure of isoheptane (C7H16) is a branched alkane, not a linear one.

Correct Answer: False

Solution:

For a compound microscope to produce a magnified image, the object must be placed closer than the focal point of the objective lens. Placing it at the focal point would not produce a magnified image.

Correct Answer: True

Solution:

A convex lens can form a virtual image when the object is placed between the lens and its focal point.

Correct Answer: True

Solution:

Total internal reflection occurs when light travels from a denser medium to a rarer medium and the angle of incidence exceeds the critical angle.

Correct Answer: True

Solution:

The excerpt mentions that a convex lens can form an image on a screen placed 90 cm from an object.

Correct Answer: True

Solution:

The virtual image formed by a convex mirror is always smaller than the object and appears between the focal point and the pole of the mirror.

Correct Answer: True

Solution:

A convex mirror always forms a virtual image, which is diminished and located between the focus and the pole.

Correct Answer: False

Solution:

The refractive index of a medium is not solely determined by its mass density. Optical density, which affects the refractive index, is related to the speed of light in the medium, not just its mass density.

Correct Answer: True

Solution:

The diagram description explicitly states that isoheptane is a branched alkane.

Correct Answer: True

Solution:

A concave mirror can form a real and inverted image when light rays converge at the focal point after reflection.

Correct Answer: False

Solution:

The magnification in absolute size is not always equal to the angular magnification or magnifying power of an instrument.

Correct Answer: True

Solution:

When an object is placed between the focal point and a concave mirror, the image formed is virtual, upright, and enlarged.

Correct Answer: False

Solution:

The diagram of isoheptane shows it as a branched alkane, not a linear one.

Correct Answer: True

Solution:

The refractive index is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium, which is always greater than 1.

Correct Answer: True

Solution:

According to the ray diagram description, a concave mirror reflects light rays to converge at the focal point, forming a real and inverted image.

Correct Answer: False

Solution:

A convex mirror always forms a virtual image, regardless of the object's position.

Correct Answer: False

Solution:

A convex mirror always produces a virtual image, not a real one, regardless of the object's location.

Correct Answer: True

Solution:

A concave mirror can converge light rays to form a real image at the focal point.

Correct Answer: True

Solution:

A convex lens can form a real image on a screen when the object is placed at a suitable distance from the lens.

Correct Answer: False

Solution:

A convex mirror always produces a virtual image, regardless of the object's position.

Correct Answer: True

Solution:

A convex lens converges light rays, resulting in a positive focal length.

Correct Answer: False

Solution:

The structure of isoheptane is a branched alkane, not a linear one.

Correct Answer: True

Solution:

When an object is placed between the focal point and the pole of a concave mirror, it produces a virtual, enlarged image.

Correct Answer: True

Solution:

A concave mirror can form a real and inverted image that is smaller than the object when the object is placed beyond the center of curvature.

Correct Answer: True

Solution:

Total internal reflection occurs when light travels from an optically denser medium to a rarer medium and the angle of incidence exceeds the critical angle.

Correct Answer: True

Solution:

When an object is placed between the pole and the focus of a concave mirror, it produces a virtual and enlarged image.

Correct Answer: False

Solution:

A concave mirror can produce a real image when the object is placed beyond its focal point.

Correct Answer: False

Solution:

The description clarifies that optical density (refractive index) should not be confused with mass density, which is mass per unit volume.

Correct Answer: True

Solution:

A concave mirror can produce an inverted image when the object is placed beyond the focal point.

Correct Answer: True

Solution:

Concave mirrors converge light rays at the focal point after reflection, as demonstrated by ray diagrams.

Correct Answer: True

Solution:

When an object is placed closer to a convex lens than its focal length, a virtual image is formed on the same side as the object.

Correct Answer: True

Solution:

When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect, and enlarged.

Correct Answer: False

Solution:

Isoheptane is a branched alkane, not a straight-chain alkane.

Correct Answer: True

Solution:

A concave mirror forms a real image beyond the center of curvature when an object is placed between the focus and the center of curvature.

Correct Answer: False

Solution:

Total internal reflection occurs if the incident angle is greater than the critical angle.

Correct Answer: False

Solution:

A concave mirror can form both real and virtual images depending on the position of the object relative to the focal point.

Correct Answer: False

Solution:

A convex mirror always produces a diminished image, not an enlarged one.

Correct Answer: False

Solution:

Isoheptane is a branched alkane, not a linear one.

Correct Answer: False

Solution:

A magnifying glass, which is a converging lens, typically produces a virtual image that is larger than the object.

Correct Answer: True

Solution:

The refractive index affects how much a light ray bends when it enters a new medium, according to Snell's Law.

Correct Answer: False

Solution:

Optical density (refractive index) should not be confused with mass density. A medium can have a higher optical density but lower mass density compared to another medium.

Correct Answer: True

Solution:

The lateral shift of a light ray as it passes through a medium is influenced by the thickness of the medium and the angles of incidence and refraction.

Correct Answer: True

Solution:

A compound microscope consists of an objective lens and an eyepiece lens. The objective lens forms an enlarged image of the object, which is further magnified by the eyepiece.

Correct Answer: True

Solution:

The description of refraction through a glass slab shows that the apparent depth is the real depth divided by the refractive index, indicating it is less than the real depth.

Correct Answer: False

Solution:

Optical density, indicated by the refractive index, is different from mass density. Optical density relates to the speed of light in the medium, not its mass per unit volume.

Correct Answer: True

Solution:

A convex lens can form a real image on a screen if the distance between the lens and the screen is greater than the focal length of the lens.

Correct Answer: True

Solution:

A convex lens can produce a virtual image that is larger than the object when the object is placed between the focal point and the lens. This is a common setup for magnifying glasses.

Correct Answer: False

Solution:

The refractive index of a medium is always greater than or equal to 1, as it is the ratio of the speed of light in a vacuum to the speed of light in the medium.

Correct Answer: True

Solution:

The magnifying power of a compound microscope depends on the focal lengths of the lenses and their separation.

Correct Answer: True

Solution:

The refractive index is indeed defined as the ratio of the speed of light in a vacuum to its speed in the medium, which determines how much the light bends when entering the medium.

Correct Answer: False

Solution:

A convex mirror always forms a virtual image, regardless of the position of the object.

Correct Answer: False

Solution:

When light travels from a denser to a rarer medium, the angle of refraction is greater than the angle of incidence.

Correct Answer: False

Solution:

When an object is placed between the focal point and a concave mirror, the image formed is virtual and enlarged.

Correct Answer: False

Solution:

A convex mirror always produces a virtual image regardless of the object's position.

Correct Answer: True

Solution:

The refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium.

Correct Answer: True

Solution:

A convex lens can form a real image on a screen when the object is placed at a distance greater than twice the focal length. This is because the lens focuses light rays to converge at a point on the opposite side.

Correct Answer: False

Solution:

Optical density (refractive index) is not directly related to mass density. A medium can have a higher optical density but a lower mass density compared to another medium.

Correct Answer: True

Solution:

A convex lens can form images at two different locations on a screen due to its ability to focus light from different distances, as described in the exercise where the screen is placed 90 cm from the object.

Correct Answer: True

Solution:

A convex lens can magnify objects by creating a larger virtual image when the object is placed within the focal length.

Correct Answer: True

Solution:

When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, enlarged, and upright.

Correct Answer: True

Solution:

A compound microscope consists of an objective lens and an eyepiece lens to achieve magnification.

Correct Answer: True

Solution:

A convex lens can converge light rays to form a real image on a screen, depending on the object's position relative to the lens.

Correct Answer: True

Solution:

The focal length of a lens system can be calculated using the lens formula, considering the focal lengths of individual lenses.

Correct Answer: True

Solution:

The refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium, and it is always greater than 1.

Correct Answer: False

Solution:

The structure of isoheptane is a branched alkane, not a straight-chain alkane.

Correct Answer: False

Solution:

To form an image on the opposite wall 3m away, the focal length of the convex lens must be less than or equal to 3m. A focal length greater than 3m would not converge the light rays at the required distance.

Correct Answer: True

Solution:

A convex lens can form a real image when the object is placed beyond the focal point and a virtual image when the object is placed between the lens and its focal point.

Ray Optics and Optical Instruments

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

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Chapter Nine: Ray Optics and Optical Instruments

9.1 Introduction

9.2 Key Concepts

9.3 Exercises

9.1

9.2

9.3

9.4

9.5

9.4 Important Formulas

9.5 Common Mistakes and Exam Tips

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Ray Optics and Optical Instruments

Common Pitfalls

Tips for Success

Multiple Choice Questions

Q1.A card sheet divided into squares each of size 1 mm² is viewed at a distance of 9 cm through a magnifying glass of focal length 9 cm. What is the magnification produced by the lens?

Correct Answer: A

Q2.What is the maximum possible focal length of a convex lens needed to project an image of a bulb on a wall 3m away?

Correct Answer: A

Q3.A concave mirror has its focal point at 10 cm from the mirror. An object is placed 15 cm from the mirror. What type of image is formed?

Correct Answer: A

Q4.A glass slab of thickness 15 cm and refractive index 1.5 is placed on a table. By what distance would a pin appear to be raised if viewed from above?

Correct Answer: A

Q5.In a ray diagram, a concave mirror forms an image of an object placed at a distance of 30 cm from the mirror. If the focal length of the mirror is 10 cm, where is the image formed?

Correct Answer: B

Q6.What is the magnification produced by a magnifying glass with a focal length of 9 cm when viewing an object at a distance of 9 cm?

Correct Answer: B

Q7.What is the maximum possible focal length of a convex lens required to obtain the image of a small electric bulb fixed on the wall of a room on the opposite wall 3 meters away?

Correct Answer: A

Q8.What happens to light rays when they pass from a denser medium to a rarer medium?

Correct Answer: B

Q9.What is the effect of a glass slab on the apparent position of an object viewed through it?

Correct Answer: A

Q10.In a ray diagram of a concave mirror, if an object is placed between the focal point (FFF) and the pole (PPP) of the mirror, what type of image is formed?

Correct Answer: B

Q11.What is the critical angle for a light ray traveling from glass (refractive index 1.5) to air?

Correct Answer: A

Q12.A light ray enters a rectangular glass slab with an angle of incidence of 30°. If the refractive index of the glass is 1.5, what is the angle of refraction inside the glass?

Correct Answer: A

Q13.A ray of light passes from a denser medium to a rarer medium. What happens to the angle of refraction compared to the angle of incidence?

Correct Answer: C

Q14.A ray of light travels from glass (refractive index 1.5) into air (refractive index 1.0). What is the critical angle for total internal reflection at the glass-air interface?

Correct Answer: A

Q15.Which of the following statements is true for a convex mirror?

Correct Answer: B

Q16.In a compound microscope, the objective lens has a focal length of 8.0 mm and the eyepiece has a focal length of 2.5 cm. If the object is placed 9.0 mm from the objective, what is the magnifying power of the microscope?

Correct Answer: A

Q17.What is the chemical formula of isoheptane?

Correct Answer: A

Q18.A convex lens is used to project an image of an object onto a screen 3 meters away. What is the maximum possible focal length of the lens required for this purpose?

Correct Answer: A

Q19.A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. What is the focal length of the lens?

Correct Answer: B

Q20.At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Correct Answer: B

Q21.A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope?

Correct Answer: B

Q22.A ray of light enters a rectangular glass slab from air at an angle of incidence i1i_1i1​. If the refractive index of glass is 1.5, what is the angle of refraction r1r_1r1​ inside the glass?

Correct Answer: A

Q23.A convex lens is used to project the image of a small electric bulb onto a screen placed 3 meters away. If the maximum possible focal length of the lens required for this purpose is to be calculated, what is the value?

Correct Answer: A

Q24.What is the angular magnification of a telescope with an objective lens of focal length 15 m and an eyepiece of focal length 1 cm?

Correct Answer: A

Q25.A concave mirror has a focal length of 15 cm. An object is placed 10 cm from the mirror. What type of image is formed?

Correct Answer: B

Q26.What is the angular magnification of a telescope with an objective lens of focal length 15 m and an eyepiece of focal length 1 cm?

Correct Answer: A

Q27.A ray of light is incident on a prism with a refracting angle of 60°, and it just suffers total internal reflection at the other face. Given the refractive index of the prism material is 1.524, what is the angle of incidence?

Correct Answer: C

Q28.What happens to a light ray when it enters a denser medium at an angle?

Correct Answer: A

Q29.A ray of light enters a rectangular glass slab with an angle of incidence i1i_1i1​ and emerges with an angle of emergence r2r_2r2​. If the refractive index of the glass is 1.5, what is the lateral shift of the ray if the thickness of the slab is 5 cm?

Correct Answer: B

Q30.A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision (25 cm)?

Correct Answer: A