Question 1

What is the phenomenon called when light splits into its constituent colors due to wavelength-dependent refractive index?

Accepted Answer

Correct Option: C. Solution: Dispersion is the splitting of light into its constituent colors due to wavelength-dependent refractive index.

Question 2

A simple microscope uses a convex lens with a small focal length. If the focal length of the lens is 5 cm, what is the magnifying power when the final image is at the near point (25 cm)?

Accepted Answer

Correct Option: C. Solution: The magnifying power $m$ for a simple microscope with the image at the near point is given by $m = 1 + \frac{D}{f}$, where $D = 25 	ext{ cm}$ and $f = 5 	ext{ cm}$. Thus, $m = 1 + \frac{25}{5} = 6$.

Question 3

When light passes from a denser medium to a rarer medium, under what condition does total internal reflection occur?

Accepted Answer

Correct Option: C. Solution: Total internal reflection occurs when the angle of incidence exceeds the critical angle for the given pair of media.

Question 4

If a simple microscope has a focal length of 5 cm, what is the magnifying power when the image is at infinity?

Accepted Answer

Correct Option: A. Solution: For a simple microscope with the image at infinity, the magnifying power is $m = \frac{D}{f}$. Given $D = 25$ cm and $f = 5$ cm, $m = \frac{25}{5} = 5$.

Question 5

What is the approximate magnifying power of a compound microscope when the final image is at infinity?

Accepted Answer

Correct Option: A. Solution: The magnifying power of a compound microscope for the final image at infinity is given by $m = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)$, where $L$ is the tube length, $f_o$ is the focal length of the objective, and $f_e$ is the focal length of the eyepiece.

Question 6

A convex lens with a focal length of 10 cm is used to form an image of an object placed 15 cm away from it. What is the nature and position of the image formed?

Accepted Answer

Correct Option: A. Solution: Using the lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$, where $u = -15$ cm and $f = 10$ cm, we find $v = 30$ cm. The image is real and inverted.

Question 7

A compound microscope consists of an objective lens with a focal length of $f_o = 1 	ext{ cm}$ and an eyepiece with a focal length of $f_e = 5 	ext{ cm}$. If the tube length is $L = 20 	ext{ cm}$, what is the total magnification of the microscope?

Accepted Answer

Correct Option: A. Solution: The total magnification $M$ of a compound microscope is given by the product of the magnifications of the objective and the eyepiece: $M = m_o 	imes m_e$, where $m_o = \frac{L}{f_o}$ and $m_e = \frac{D}{f_e}$. Substituting the given values, $m_o = \frac{20}{1} = 20$ and $m_e = \frac{25}{5} = 5$. Therefore, $M = 20 	imes 5 = 100$.

Question 8

In a compound microscope, the objective lens has a focal length $f_o$ and the eyepiece has a focal length $f_e$. If the tube length is $L$, what is the total magnification $M$ of the microscope when the final image is at infinity?

Accepted Answer

Correct Option: A. Solution: The total magnification of a compound microscope is given by the product of the magnification of the objective lens and the eyepiece. When the final image is at infinity, $M = \frac{L}{f_o} \cdot \frac{D}{f_e}$, where $L$ is the tube length and $D$ is the least distance of distinct vision.

Question 9

When light passes through a parallel-sided glass slab, what happens to the emergent ray?

Accepted Answer

Correct Option: B. Solution: A parallel-sided slab produces lateral shift without net deviation, meaning the emergent ray is parallel to the incident ray but shifted laterally.

Question 10

In a simple microscope, if the focal length of the lens is $5\, 	ext{cm}$, what is the angular magnification when the image is formed at the near point?

Accepted Answer

Correct Option: B. Solution: The angular magnification $m$ for a simple microscope when the image is at the near point is given by $m = 1 + \frac{D}{f}$, where $D = 25\, 	ext{cm}$ is the least distance of distinct vision. Substituting $f = 5\, 	ext{cm}$, we get $m = 1 + \frac{25}{5} = 6$.

Question 11

A mirage is an optical illusion caused by which of the following?

Accepted Answer

Correct Option: B. Solution: A mirage occurs due to total internal reflection, where light bends as it passes through layers of air at different temperatures, creating the illusion of water or a distant object.

Question 12

What is the lens formula for a thin lens?

Accepted Answer

Correct Option: B. Solution: The lens formula for a thin lens is given by $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$, where $v$ is the image distance, $u$ is the object distance, and $f$ is the focal length.

Question 13

A convex lens has a focal length of 20 cm in air. What will be its focal length when it is submerged in water with a refractive index of 1.33?

Accepted Answer

Correct Option: 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.

Question 14

A lens is designed using the Lens Maker's Formula with $n_2 = 1.5$, $n_1 = 1.0$, $R_1 = 20 	ext{ cm}$, and $R_2 = -30 	ext{ cm}$. What is the focal length of the lens?

Accepted Answer

Correct Option: B. Solution: Using the Lens Maker's Formula: $$\frac{1}{f} = \left(\frac{1.5}{1.0} - 1ight)\left(\frac{1}{20} - \frac{1}{-30}ight)$$, we find that $f = 15 	ext{ cm}$.

Question 15

For refraction from medium $n_1$ to $n_2$ at a spherical surface with radius $R$, which equation describes the image formation using the Cartesian sign convention?

Accepted Answer

Correct Option: A. Solution: The correct equation for refraction at a spherical surface using the Cartesian sign convention is $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$.

Question 16

In a diamond, the sparkle is primarily due to which optical phenomenon?

Accepted Answer

Correct Option: B. Solution: The sparkle of a diamond is due to total internal reflection, which occurs because of the high refractive index of diamond, causing light to be trapped and reflected multiple times within the gemstone.

Question 17

According to the Cartesian sign convention in ray optics, how are distances measured along the direction of incident light?

Accepted Answer

Correct Option: A. Solution: In the Cartesian sign convention, distances measured along the direction of incident light are considered positive.

Question 18

For a lens with focal length $f = 10 	ext{ cm}$, what is the power of the lens in dioptres?

Accepted Answer

Correct Option: C. Solution: The power $P$ of a lens is given by $$P = \frac{1}{f}$$, where $f$ is in meters. For $f = 10 	ext{ cm} = 0.1 	ext{ m}$, the power is $$P = \frac{1}{0.1} = 10 	ext{ D}$$.

Question 19

In a compound microscope, what is the role of the objective lens?

Accepted Answer

Correct Option: A. Solution: In a compound microscope, the objective lens forms a real, inverted, and magnified image of the object, which then serves as the object for the eyepiece.

Question 20

In a prism, if the angle of incidence is equal to the angle of emergence and the angle of deviation is minimum, what can be inferred about the angles of refraction inside the prism?

Accepted Answer

Correct Option: A. Solution: At minimum deviation, the path of the light ray inside the prism is symmetrical, meaning the angles of refraction are equal.

Question 21

In an optical fibre, what is the primary reason that allows light to be transmitted over long distances with minimal loss?

Accepted Answer

Correct Option: B. Solution: Optical fibres transmit light efficiently due to repeated total internal reflection, which keeps the light within the core of the fibre.

Question 22

A point source of light is placed 100 cm away from a spherical glass surface with a refractive index of 1.5 and a radius of curvature of 20 cm. Where will the image be formed?

Accepted Answer

Correct Option: B. Solution: Using the formula for refraction at a spherical surface: $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ where $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm, and $R = 20$ cm. Solving gives $v = 100$ cm.

Question 23

A parallel-sided glass slab with a refractive index of 1.5 is placed over a printed page. The thickness of the slab is 6 cm. What is the lateral shift of the print when viewed from directly above?

Accepted Answer

Correct Option: C. Solution: The lateral shift $d$ is given by the formula $d = t \left(1 - \frac{1}{n}\right)$ where $t$ is the thickness of the slab and $n$ is the refractive index. Substituting $t = 6$ cm and $n = 1.5$, we get $d = 6 \left(1 - \frac{1}{1.5}\right) = 2$ cm.

Question 24

What is the mirror equation used for spherical mirrors?

Accepted Answer

Correct Option: A. Solution: The mirror equation for spherical mirrors is $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$, where $v$ is the image distance, $u$ is the object distance, and $f$ is the focal length.

Question 25

A convex lens has a focal length of 10 cm. An object is placed 20 cm from the lens. What is the magnification of the image formed?

Accepted Answer

Correct Option: A. Solution: Using the lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ and solving for $v$, we find $v = 20$ cm. Magnification $m = \frac{v}{u} = 0.5$. The image is real and inverted.

Question 26

A compound microscope has an objective lens with a focal length of 2 cm and an eyepiece with a focal length of 5 cm. If the tube length is 20 cm, what is the total magnification of the microscope?

Accepted Answer

Correct Option: A. Solution: The magnification of a compound microscope is given by $m = m_o 	imes m_e$, where $m_o = \frac{L}{f_o}$ and $m_e = \frac{D}{f_e}$. Substituting $L = 20$ cm, $f_o = 2$ cm, $f_e = 5$ cm, and $D = 25$ cm, we find $m_o = 10$ and $m_e = 2.5$. Thus, $m = 10 	imes 2.5 = 25$.

Question 27

A concave mirror has a radius of curvature of 30 cm. What is its focal length?

Accepted Answer

Correct Option: A. Solution: The focal length $f$ of a concave mirror is given by $f = R/2$. Therefore, for a radius of curvature $R = 30$ cm, the focal length $f = 30/2 = 15$ cm.

Question 28

Which of the following applications utilizes the principle of total internal reflection?

Accepted Answer

Correct Option: B. Solution: Optical fibers guide light through repeated total internal reflection due to a higher refractive index core and lower refractive index cladding.

Question 29

What is the relationship between the focal length $f$ and the radius of curvature $R$ for a spherical mirror?

Accepted Answer

Correct Option: B. Solution: For spherical mirrors, the focal length $f$ is half the radius of curvature $R$, hence $f = R/2$.

Question 30

When using a compound microscope, the objective lens forms an intermediate image that is:

Accepted Answer

Correct Option: A. Solution: In a compound microscope, the objective lens forms a real, inverted, and magnified image of the object. This intermediate image is then further magnified by the eyepiece to produce the final image.

Question 31

In an optical fiber, what causes the light to be guided through the core?

Accepted Answer

Correct Option: B. Solution: In optical fibers, the core has a higher refractive index than the cladding, causing light to be guided through repeated total internal reflection.

Question 32

What is the primary function of the objective lens in a compound microscope?

Accepted Answer

Correct Option: A. Solution: The objective lens in a compound microscope forms a real, inverted, and magnified image of the object, which then serves as the object for the eyepiece.

Question 33

In a concave mirror setup, if the object is placed at a distance of 10 cm in front of the mirror with a focal length of 7.5 cm, what is the nature of the image formed?

Accepted Answer

Correct Option: A. Solution: Using the mirror equation $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$, where $u = -10 	ext{ cm}$ and $f = -7.5 	ext{ cm}$, we find $v = 30 	ext{ cm}$. The image is real and inverted.

Question 34

In a prism, if the angle of incidence is equal to the angle of emergence, what can be inferred about the angle of deviation?

Accepted Answer

Correct Option: C. Solution: When the angle of incidence equals the angle of emergence, the angle of deviation is at its minimum value.

Question 35

What is the magnifying power of a simple microscope when the final image is at the near point?

Accepted Answer

Correct Option: A. Solution: The magnifying power of a simple microscope when the final image is at the near point is given by $m = 1 + \frac{D}{f}$, where $D$ is the least distance of distinct vision and $f$ is the focal length of the lens.

Question 36

If the object is placed at the focal point of the objective lens in a compound microscope, where is the first image formed?

Accepted Answer

Correct Option: A. Solution: When the object is placed at the focal point of the objective lens, the rays emerge parallel and the first image is formed at infinity.

Question 37

A concave mirror forms a real image of an object placed 10 cm in front of it. If the radius of curvature of the mirror is 15 cm, where is the image formed?

Accepted Answer

Correct Option: C. Solution: Using the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, where $f = \frac{R}{2} = 7.5$ cm and $u = -10$ cm, we find $v = 30$ cm in front of the mirror.

Question 38

A point source of light is placed 100 cm away from a spherical glass surface with a refractive index of 1.5 and a radius of curvature of 20 cm. Where will the image be formed?

Accepted Answer

Correct Option: B. Solution: Using the formula $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ with $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm, and $R = 20$ cm, we find that the image is formed 100 cm from the surface.

Question 39

What condition must be met for total internal reflection to occur?

Accepted Answer

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

Question 40

A beam of white light is incident on a prism at an angle of incidence $i$. If the angle of deviation $	heta$ is minimized, which of the following statements is true about the path of light inside the prism?

Accepted Answer

Correct Option: A. Solution: At the minimum angle of deviation, the refracted ray inside the prism is parallel to the base of the prism, which results in the angles of incidence and emergence being equal.

Question 41

A light ray travels from air into a glass medium. According to the Cartesian sign convention, what is the sign of the image distance if the image forms on the same side as the incident light?

Accepted Answer

Correct Option: B. Solution: According to the Cartesian sign convention, distances measured in the direction opposite to the incident light are negative. Thus, the image distance is negative if the image forms on the same side as the incident light.

Question 42

A beam of light passes through a prism with an angle of minimum deviation $D_m$. If the angle of the prism is $A$ and the refractive index is $n_{21}$, which of the following equations correctly represents the relationship between these variables?

Accepted Answer

Correct Option: A. Solution: At minimum deviation, the refractive index is given by $n_{21} = \frac{\sin((A + D_m)/2)}{\sin(A/2)}$, which is derived from the geometry of the prism.

Question 43

If the critical angle for a medium is 48.75°, what is the refractive index of the medium with respect to air?

Accepted Answer

Correct Option: A. Solution: The refractive index of water with respect to air is 1.33, corresponding to a critical angle of 48.75°.

Question 44

What is the power of a lens with a focal length of 0.5 meters?

Accepted Answer

Correct Option: A. Solution: The power of a lens is given by $P = \frac{1}{f}$, where $f$ is the focal length in meters. For $f = 0.5$ meters, $P = \frac{1}{0.5} = 2$ dioptres.

Question 45

Which of the following conditions is necessary for total internal reflection to occur in an optical fibre?

Accepted Answer

Correct Option: B. Solution: For total internal reflection to occur, the refractive index of the core must be greater than that of the cladding, allowing light to be reflected back into the core.

Question 46

If an object is placed 10 cm in front of a concave mirror with a focal length of 5 cm, where will the image be formed?

Accepted Answer

Correct Option: B. Solution: Using the mirror equation $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$, with $u = -10$ cm and $f = -5$ cm (concave mirror), we find $v = -10$ cm, meaning the image is formed 10 cm in front of the mirror.

Question 47

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

Accepted Answer

Correct Option: B. Solution: Using Snell's Law, $n_1 \sin i = n_2 \sin r$, where $n_1 = 1$, $n_2 = 1.5$, and $i = 30°$, we find $\sin r = \frac{1}{1.5} \sin 30° = 0.333$. Thus, $r = \arcsin(0.333) \approx 20.49°$.

Question 48

If a light ray passes from air into water, which statement about the refracted ray is correct given that the refractive index of water is greater than that of air?

Accepted Answer

Correct Option: A. Solution: When light passes from a medium of lower refractive index (air) to a medium of higher refractive index (water), the refracted ray bends towards the normal.

Question 49

A convex lens has a focal length of 10 cm. If an object is placed 15 cm from the lens, where will the image be formed?

Accepted Answer

Correct Option: B. Solution: Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, where $f = 10$ cm and $u = -15$ cm, we solve for $v$: $\frac{1}{v} = \frac{1}{10} + \frac{1}{15} = \frac{5}{30}$. Thus, $v = 30$ cm, indicating the image is formed 30 cm on the opposite side of the lens.

Question 50

Which of the following statements best describes the principle of total internal reflection as utilized in optical fibers?

Accepted Answer

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

Question 51

In a prism, what is the relationship between the angles of incidence ($i$), emergence ($e$), and the angle of deviation ($\delta$)?

Accepted Answer

Correct Option: A. Solution: The angle of deviation $\delta$ in a prism is given by the equation $\delta = i + e - A$, where $A$ is the angle of the prism.

Question 52

At minimum deviation in a prism, which of the following conditions hold true?

Accepted Answer

Correct Option: A. Solution: At minimum deviation, the angles of incidence and emergence are equal ($i = e$), and the angles of refraction are equal to half the angle of the prism ($r_1 = r_2 = \frac{A}{2}$).

Question 53

A ray of light is incident on a prism at an angle of 45°. If the prism has an angle of 60° and the refractive index is 1.6, what is the angle of deviation?

Accepted Answer

Correct Option: B. Solution: Using the formula for deviation in a prism, $\delta = i + e - A$, and knowing $r_1 + r_2 = A$, we find $r_1 = \arcsin(\frac{\sin 45°}{1.6})$ and $r_2 = A - r_1$. Finally, $\delta = 45° + \arcsin(1.6 \sin r_2) - 60° \approx 38.68°$.

Question 54

Consider a spherical glass surface with a refractive index of 1.5 and a radius of curvature of 20 cm. If light from a point source in air is incident on this surface from a distance of 100 cm, at what distance from the glass surface will the image be formed?

Accepted Answer

Correct Option: B. Solution: Using the formula $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ with $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm, and $R = 20$ cm, we find $v = 100$ cm.

Question 55

A ray of light passes through a glass prism with an angle of incidence of 30°. If the angle of deviation is 40°, what is the angle of emergence?

Accepted Answer

Correct Option: C. Solution: Using the formula for deviation, $$\delta = i + e - A$$, where $\delta$ is the deviation, $i$ is the angle of incidence, $e$ is the angle of emergence, and $A$ is the angle of the prism. Rearranging, we find $e = \delta + A - i$. Given $\delta = 40°$, $i = 30°$, and assuming a typical prism angle $A = 60°$, we solve for $e$: $$e = 40° + 60° - 30° = 70°$$. Therefore, the angle of emergence is 70°.

Question 56

What is the Lens Maker's Formula used for?

Accepted Answer

Correct Option: A. Solution: The Lens Maker's Formula relates the focal length of a lens to the refractive indices and radii of curvature: $$\frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$.

Question 57

In a ray diagram for a convex lens, which ray emerges parallel to the principal axis after passing through the lens?

Accepted Answer

Correct Option: C. Solution: A ray passing through the first principal focus of a convex lens emerges parallel to the principal axis after refraction.

Question 58

A light ray is incident on a concave mirror parallel to its principal axis. After reflection, it passes through a point on the principal axis. Identify this point.

Accepted Answer

Correct Option: B. Solution: For a concave mirror, a ray parallel to the principal axis passes through the principal focus after reflection.

Question 59

Which of the following is true about the critical angle for total internal reflection?

Accepted Answer

Correct Option: 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°.

Question 60

Which of the following statements about a simple microscope is incorrect?

Accepted Answer

Correct Option: B. Solution: A simple microscope forms a virtual and erect image, not a real and inverted one. Therefore, option B is incorrect.

Question 61

Which of the following statements is true about the final image formed by a compound microscope?

Accepted Answer

Correct Option: C. Solution: In a compound microscope, the final image formed by the eyepiece is virtual and erect with respect to the original object.

Question 62

A convex lens is used to form a real image of an object placed $30$ cm from the lens. If the image is formed $60$ cm from the lens on the opposite side, what is the focal length of the lens?

Accepted Answer

Correct Option: B. Solution: Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, where $v = 60$ cm and $u = -30$ cm (since object distance is taken as negative), we find $\frac{1}{f} = \frac{1}{60} - \left(-\frac{1}{30}\right) = \frac{1}{60} + \frac{1}{30} = \frac{1}{20}$. Thus, the focal length $f = 20$ cm.

Question 63

When light passes through a prism at minimum deviation, the angle of incidence $i$ is equal to the angle of emergence $e$. If the refractive index is $n_{21}$ and the prism angle is $A$, what is the expression for the angle of minimum deviation $D_m$?

Accepted Answer

Correct Option: A. Solution: At minimum deviation, $i = e$ and the deviation $D_m = i + e - A = 2i - A$.

Question 64

What principle allows optical fibers to transmit light with minimal loss?

Accepted Answer

Correct Option: C. Solution: Optical fibers utilize the principle of total internal reflection to guide light through the fiber with minimal loss.

Ray Optics and Optical In..

Learning Objectives

Revision Notes & Summary

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

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Ray Optics and Optical Instruments

Common Pitfalls

Tips for Success

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Multiple Choice Questions

Q1.What is the phenomenon called when light splits into its constituent colors due to wavelength-dependent refractive index?

Correct Answer: C

Q2.A simple microscope uses a convex lens with a small focal length. If the focal length of the lens is 5 cm, what is the magnifying power when the final image is at the near point (25 cm)?

Correct Answer: C

Q3.When light passes from a denser medium to a rarer medium, under what condition does total internal reflection occur?

Correct Answer: C

Q4.If a simple microscope has a focal length of 5 cm, what is the magnifying power when the image is at infinity?

Correct Answer: A

Q5.What is the approximate magnifying power of a compound microscope when the final image is at infinity?

Correct Answer: A

Q6.A convex lens with a focal length of 10 cm is used to form an image of an object placed 15 cm away from it. What is the nature and position of the image formed?

Correct Answer: A

Q7.A compound microscope consists of an objective lens with a focal length of fo=1 cmf_o = 1 \text{ cm}fo​=1 cm and an eyepiece with a focal length of fe=5 cmf_e = 5 \text{ cm}fe​=5 cm. If the tube length is L=20 cmL = 20 \text{ cm}L=20 cm, what is the total magnification of the microscope?

Correct Answer: A

Q8.In a compound microscope, the objective lens has a focal length fof_ofo​ and the eyepiece has a focal length fef_efe​. If the tube length is LLL, what is the total magnification MMM of the microscope when the final image is at infinity?

Correct Answer: A

Q9.When light passes through a parallel-sided glass slab, what happens to the emergent ray?

Correct Answer: B

Q10.In a simple microscope, if the focal length of the lens is 5 cm5\, \text{cm}5cm, what is the angular magnification when the image is formed at the near point?

Correct Answer: B

Q11.A mirage is an optical illusion caused by which of the following?

Correct Answer: B

Q12.What is the lens formula for a thin lens?

Correct Answer: B

Q13.A convex lens has a focal length of 20 cm in air. What will be its focal length when it is submerged in water with a refractive index of 1.33?

Correct Answer: D

Q14.A lens is designed using the Lens Maker's Formula with n2=1.5n_2 = 1.5n2​=1.5, n1=1.0n_1 = 1.0n1​=1.0, R1=20 cmR_1 = 20 \text{ cm}R1​=20 cm, and R2=−30 cmR_2 = -30 \text{ cm}R2​=−30 cm. What is the focal length of the lens?

Correct Answer: B

Q15.For refraction from medium n1n_1n1​ to n2n_2n2​ at a spherical surface with radius RRR, which equation describes the image formation using the Cartesian sign convention?

Correct Answer: A

Q16.In a diamond, the sparkle is primarily due to which optical phenomenon?

Correct Answer: B

Q17.According to the Cartesian sign convention in ray optics, how are distances measured along the direction of incident light?

Correct Answer: A

Q18.For a lens with focal length f=10 cmf = 10 \text{ cm}f=10 cm, what is the power of the lens in dioptres?

Correct Answer: C

Q19.In a compound microscope, what is the role of the objective lens?

Correct Answer: A

Q20.In a prism, if the angle of incidence is equal to the angle of emergence and the angle of deviation is minimum, what can be inferred about the angles of refraction inside the prism?

Correct Answer: A

Q21.In an optical fibre, what is the primary reason that allows light to be transmitted over long distances with minimal loss?

Correct Answer: B

Q22.A point source of light is placed 100 cm away from a spherical glass surface with a refractive index of 1.5 and a radius of curvature of 20 cm. Where will the image be formed?

Correct Answer: B

Q23.A parallel-sided glass slab with a refractive index of 1.5 is placed over a printed page. The thickness of the slab is 6 cm. What is the lateral shift of the print when viewed from directly above?

Correct Answer: C

Q24.What is the mirror equation used for spherical mirrors?

Correct Answer: A

Q25.A convex lens has a focal length of 10 cm. An object is placed 20 cm from the lens. What is the magnification of the image formed?

Correct Answer: A

Q26.A compound microscope has an objective lens with a focal length of 2 cm and an eyepiece with a focal length of 5 cm. If the tube length is 20 cm, what is the total magnification of the microscope?

Correct Answer: A

Q27.A concave mirror has a radius of curvature of 30 cm. What is its focal length?

Correct Answer: A