Ray Optics Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A telescope has an objective of focal length \( 120 \, \text{cm} \) and an eyepiece of focal length \( 6 \, \text{cm} \). What is its magnifying power?

Magnifying power: \( m = \frac{f_o}{f_e} \).

\( f_o = 120 \, \text{cm} \), \( f_e = 6 \, \text{cm} \).

\( m = \frac{120}{6} = 20 \).

15
18
20
25
3

A converging beam meets a convex lens (\( f = 20 \, \text{cm} \)) \( 8 \, \text{cm} \) before the convergence point. What is the new image distance?

Object distance: \( u = -8 \, \text{cm} \) (virtual object).

Focal length: \( f = 20 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{-8} = \frac{1}{20} \Rightarrow \frac{1}{v} + \frac{1}{8} = \frac{1}{20} \).

\( \frac{1}{v} = \frac{1}{20} - \frac{1}{8} = \frac{2 - 5}{40} = \frac{-3}{40} \).

\( v = -\frac{40}{3} \approx -13.33 \, \text{cm} \) (13.33 cm to the left).

10 cm
12 cm
14 cm
13.3 cm
4

A double convex lens of refractive index \( 1.5 \) has radii of curvature \( 18 \, \text{cm} \) and \( -18 \, \text{cm} \). What is its focal length?

Lens maker’s formula: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).

\( n = 1.5 \), \( R_1 = 18 \, \text{cm} \), \( R_2 = -18 \, \text{cm} \).

\( \frac{1}{f} = (1.5 - 1) \left( \frac{1}{18} - \frac{1}{-18} \right) = 0.5 \left( \frac{1}{18} + \frac{1}{18} \right) = 0.5 \times \frac{2}{18} = \frac{1}{18} \).

\( f = 18 \, \text{cm} \).

15 cm
18 cm
20 cm
25 cm
2

What is the critical angle for a crown glass (\( n = 1.52 \)) to air interface?

Critical angle: \( \sin i_c = \frac{n_2}{n_1} \).

Crown glass (\( n_1 = 1.52 \)), air (\( n_2 = 1 \)).

\( \sin i_c = \frac{1}{1.52} \approx 0.658 \).

\( i_c = \sin^{-1}(0.658) \approx 41.1^\circ \).

41°
45°
50°
35°
1

A double convex lens has radii of curvature \( 30 \, \text{cm} \) and \( -30 \, \text{cm} \) with refractive index \( 1.6 \). What is its focal length?

Lens maker’s formula: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).

\( n = 1.6 \), \( R_1 = 30 \, \text{cm} \), \( R_2 = -30 \, \text{cm} \).

\( \frac{1}{f} = (1.6 - 1) \left( \frac{1}{30} - \frac{1}{-30} \right) = 0.6 \left( \frac{1}{30} + \frac{1}{30} \right) = 0.6 \times \frac{2}{30} = \frac{1.2}{30} = \frac{1}{25} \).

\( f = 25 \, \text{cm} \).

20 cm
25 cm
30 cm
35 cm
2

Why is the image formed by a plane mirror always virtual?

In a plane mirror, the reflected rays do not actually converge but appear to diverge from a point behind the mirror when traced backward. This apparent origin of rays behind the mirror results in a virtual image that cannot be projected onto a screen.

Due to light converging in front of the mirror
Due to apparent divergence from behind the mirror
Due to the mirror’s curvature
Due to dispersion of light
2

A simple microscope uses a lens of focal length \( 8 \, \text{cm} \). What is the magnification when the image is at infinity?

Magnification at infinity: \( m = \frac{D}{f} \).

\( D = 25 \, \text{cm} \), \( f = 8 \, \text{cm} \).

\( m = \frac{25}{8} = 3.125 \).

2.5
3.1
3.5
4
2

In optical fibers, what ensures that light remains confined within the core during transmission?

In optical fibers, the core has a higher refractive index than the cladding, enabling total internal reflection. When light strikes the core-cladding boundary at an angle greater than the critical angle, it reflects back into the core, ensuring confinement and minimal loss.

Absorption by the cladding
Refraction into the cladding
Higher refractive index of core than cladding
Lower refractive index of core than cladding
3

A lens has a power of \( +3 \, \text{D} \). What is its focal length in centimeters?

Power: \( P = \frac{1}{f} \) (in meters).

\( P = +3 \, \text{D} \Rightarrow 3 = \frac{1}{f} \Rightarrow f = \frac{1}{3} \approx 0.333 \, \text{m} \approx 33.3 \, \text{cm} \).

25 cm
30 cm
33 cm
40 cm
3

An object is placed \( 18 \, \text{cm} \) from a convex mirror of focal length \( 12 \, \text{cm} \). What is the image distance?

Focal length: \( f = 12 \, \text{cm} \), \( u = -18 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-18} = \frac{1}{12} \Rightarrow \frac{1}{v} = \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36} \).

\( v = \frac{36}{5} = 7.2 \, \text{cm} \) (virtual image).

6 cm
6.5 cm
7 cm
7.2 cm
4

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