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Class 12 Science (PCM)

Physics Part 2

Ray Optics and Optical Instruments

Ray Optics and Optical Instruments

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Refracting vs Reflecting Telescopes
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Dispersion of Light
Accuracy: 20%
Total Attempted5
Total Correct1
Total Incorrect4
Difficulty Level:
High
Simple Microscope / Magnifying Glass
Accuracy: 40%
Total Attempted10
Total Correct4
Total Incorrect6
Difficulty Level:
High
Compound Microscope
Accuracy: 50%
Total Attempted12
Total Correct6
Total Incorrect6
Difficulty Level:
Medium
Optical Instruments: Microscopes and Telescopes
Accuracy: 50%
Total Attempted16
Total Correct8
Total Incorrect8
Difficulty Level:
Medium
Power of a Lens
Accuracy: 50%
Total Attempted8
Total Correct4
Total Incorrect4
Difficulty Level:
Medium
Refraction and Snell's Law
Accuracy: 53%
Total Attempted17
Total Correct9
Total Incorrect8
Difficulty Level:
Medium
Lens Maker's Formula
Accuracy: 55%
Total Attempted11
Total Correct6
Total Incorrect5
Difficulty Level:
Medium
Telescope and Angular Magnification
Accuracy: 57%
Total Attempted7
Total Correct4
Total Incorrect3
Difficulty Level:
Medium
Lens Formula and Lens Magnification
Accuracy: 60%
Total Attempted5
Total Correct3
Total Incorrect2
Difficulty Level:
Medium
Mirror Equation and Magnification
Accuracy: 64%
Total Attempted14
Total Correct9
Total Incorrect5
Difficulty Level:
Medium
Optical Fibre Principle
Accuracy: 65%
Total Attempted17
Total Correct11
Total Incorrect6
Difficulty Level:
Medium
Refraction at Spherical Surfaces
Accuracy: 67%
Total Attempted9
Total Correct6
Total Incorrect3
Difficulty Level:
Medium
Combination of Lenses
Accuracy: 70%
Total Attempted10
Total Correct7
Total Incorrect3
Difficulty Level:
Medium
Refraction Through a Prism
Accuracy: 72%
Total Attempted18
Total Correct13
Total Incorrect5
Difficulty Level:
Medium
Plane Interface Refraction: Lateral Shift and Apparent Depth
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
Ray Optics and Optical Instruments
Accuracy: 75%
Total Attempted8
Total Correct6
Total Incorrect2
Difficulty Level:
Low
Total Internal Reflection
Accuracy: 75%
Total Attempted16
Total Correct12
Total Incorrect4
Difficulty Level:
Low
Reflection of Light by Spherical Mirrors
Accuracy: 80%
Total Attempted10
Total Correct8
Total Incorrect2
Difficulty Level:
Low
Focal Length of Spherical Mirrors
Accuracy: 86%
Total Attempted14
Total Correct12
Total Incorrect2
Difficulty Level:
Low
Ray Diagrams and Image Formation
Accuracy: 86%
Total Attempted7
Total Correct6
Total Incorrect1
Difficulty Level:
Low
Applications of Total Internal Reflection
Accuracy: 100%
Total Attempted9
Total Correct9
Total Incorrect0
Difficulty Level:
Low
Cartesian Sign Convention in Ray Optics
Accuracy: 100%
Total Attempted4
Total Correct4
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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  • 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.

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Revision Notes & Summary

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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.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

15 cm

B.

20 cm

C.

30 cm

D.

78.2 cm
Correct Answer: D

Solution:

Using the lens maker's formula and adjusting for the refractive indices of air (1) and water (1.33), the focal length in water is calculated to be 78.2 cm.

Chapter Concept:

Refraction at Spherical Surfaces

A.

Light is absorbed by the core and re-emitted by the cladding.

B.

Light is refracted multiple times within the core, eventually exiting at the cladding.

C.

Light undergoes repeated total internal reflection within the core due to a higher refractive index compared to the cladding.

D.

Light is reflected off the outer surface of the fiber, enhancing signal strength.
Correct Answer: C

Solution:

Optical fibers guide light through repeated total internal reflection due to the core having a higher refractive index than the cladding.

Chapter Concept:

Applications of Total Internal Reflection

A.

f=10f = 10 cm

B.

f=20f = 20 cm

C.

f=40f = 40 cm

D.

f=5f = 5 cm
Correct Answer: A

Solution:

The focal length ff of a spherical mirror is given by f=R2f = \frac{R}{2}. Therefore, f=202=10f = \frac{20}{2} = 10 cm.

Chapter Concept:

Reflection of Light by Spherical Mirrors

A.

i=ei = e and r1=r2=A2r_1 = r_2 = \frac{A}{2}

B.

i=ei = e and r1=r2=Ar_1 = r_2 = A

C.

i≠ei \neq e and r1=r2=A2r_1 = r_2 = \frac{A}{2}

D.

i≠ei \neq e and r1=r2=Ar_1 = r_2 = A
Correct Answer: A

Solution:

At minimum deviation, the angles of incidence and emergence are equal (i=ei = e), and the angles of refraction are equal to half the angle of the prism (r1=r2=A2r_1 = r_2 = \frac{A}{2}).

Chapter Concept:

Refraction Through a Prism

A.

The focal length of the eyepiece

B.

The focal length of the objective

C.

The aperture of the telescope

D.

The tube length of the telescope
Correct Answer: B

Solution:

fof_o represents the focal length of the objective lens in the telescope.

Chapter Concept:

Telescope and Angular Magnification

A.

15 cm

B.

30 cm

C.

45 cm

D.

60 cm
Correct Answer: A

Solution:

The focal length ff of a spherical mirror is given by f=R2f = \frac{R}{2}, where RR is the radius of curvature. Therefore, f=302=15f = \frac{30}{2} = 15 cm.

Chapter Concept:

Focal Length of Spherical Mirrors

A.

The angles of refraction are equal.

B.

The first angle of refraction is greater than the second.

C.

The second angle of refraction is greater than the first.

D.

The angles of refraction are complementary.
Correct Answer: A

Solution:

At minimum deviation, the path of the light ray inside the prism is symmetrical, meaning the angles of refraction are equal.

Chapter Concept:

Refraction Through a Prism

A.

It is equal to the critical angle.

B.

It is less than the critical angle.

C.

It is greater than the critical angle.

D.

It is equal to zero.
Correct Answer: C

Solution:

Total internal reflection occurs when the angle of incidence is greater than the critical angle for the pair of media.

Chapter Concept:

Optical Fibre Principle

A.

It is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

B.

It is the angle of incidence in the rarer medium for which the angle of refraction in the denser medium is 90°.

C.

It is the angle of refraction in the denser medium for which the angle of incidence in the rarer medium is zero.

D.

It is the angle of refraction in the rarer medium for which the angle of incidence in the denser medium is zero.
Correct Answer: A

Solution:

The critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

Chapter Concept:

Total Internal Reflection

A.

m=(Lfo)(Dfe)m = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)

B.

m=(Lfe)(Dfo)m = \left(\frac{L}{f_e}\right)\left(\frac{D}{f_o}\right)

C.

m=(Dfo)(Lfe)m = \left(\frac{D}{f_o}\right)\left(\frac{L}{f_e}\right)

D.

m=(foL)(feD)m = \left(\frac{f_o}{L}\right)\left(\frac{f_e}{D}\right)
Correct Answer: A

Solution:

The magnifying power of a compound microscope for the final image at infinity is given by m=(Lfo)(Dfe)m = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right), where LL is the tube length, fof_o is the focal length of the objective, and fef_e is the focal length of the eyepiece.

Chapter Concept:

Compound Microscope

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Ray Optics and Optical Instruments

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True or False

Correct Answer: False

Solution:

Snell's Law states that the refractive index depends on the pair of media and the wavelength of light, not on the angle of incidence.

Chapter Concept :

Refraction and Snell's Law

Correct Answer: True

Solution:

A parallel-sided slab causes the light ray to undergo lateral displacement, but the emergent ray is parallel to the incident ray, resulting in no net deviation.

Chapter Concept :

Plane Interface Refraction: Lateral Shift and Apparent Depth

Correct Answer: True

Solution:

The total power PP of a combination of thin lenses in contact is given by P=P1+P2+P3+...P = P_1 + P_2 + P_3 + ..., where P1,P2,P3,...P_1, P_2, P_3, ... are the powers of the individual lenses.

Chapter Concept :

Combination of Lenses

Correct Answer: True

Solution:

In spherical mirrors, the normal at any point of incidence is along the radius joining the point of incidence to the center of curvature, and the laws of reflection apply at each point.

Chapter Concept :

Reflection of Light by Spherical Mirrors

Correct Answer: True

Solution:

Both the mirror equation and the magnification formula are valid for concave and convex mirrors when the correct sign convention is used. The sign convention involves measuring distances from the pole of the mirror and assigning positive or negative signs based on the direction of the incident light.

Chapter Concept :

Mirror Equation and Magnification

Correct Answer: True

Solution:

The formula is derived using the principles of refraction and the Cartesian sign convention, which considers the direction of incident light and the geometry of the spherical surface.

Chapter Concept :

Refraction at Spherical Surfaces

Correct Answer: True

Solution:

The relationship between the focal length ff and the radius of curvature RR for spherical mirrors is derived from the geometry of reflection, where f=R2f = \frac{R}{2}.

Chapter Concept :

Focal Length of Spherical Mirrors

Correct Answer: True

Solution:

The lens maker's formula is given by 1f=(n2n1−1)(1R1−1R2)\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right), where ff is the focal length, n1n_1 and n2n_2 are the refractive indices, and R1R_1 and R2R_2 are the radii of curvature. This formula is used to design lenses with specific focal lengths.

Chapter Concept :

Lens Maker's Formula

Correct Answer: True

Solution:

The lens maker's formula, given by 1f=(n2n1−1)(1R1−1R2)\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right), is applicable to both concave and convex lenses as it relates the focal length to the refractive indices and radii of curvature.

Chapter Concept :

Lens Maker's Formula

Correct Answer: True

Solution:

Total internal reflection is a phenomenon that occurs when light travels from a denser medium to a rarer medium and the angle of incidence exceeds the critical angle, causing the light to be completely reflected back into the denser medium.

Chapter Concept :

Total Internal Reflection

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