Ray Optics Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A concave mirror of focal length \( 10 \, \text{cm} \) forms an image \( 20 \, \text{cm} \) from the mirror. What is the object distance?

Focal length: \( f = -10 \, \text{cm} \) (concave mirror).

Image distance: \( v = -20 \, \text{cm} \) (real image, same side as object).

Mirror equation: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{-20} + \frac{1}{u} = \frac{1}{-10} \Rightarrow \frac{1}{u} = \frac{1}{-10} + \frac{1}{20} = \frac{-2 + 1}{20} = \frac{-1}{20} \).

\( u = -20 \, \text{cm} \).

20 cm
15 cm
25 cm
30 cm
1

An object is placed \( 10 \, \text{cm} \) from a convex mirror of focal length \( 15 \, \text{cm} \). What is the image distance?

Focal length: \( f = 15 \, \text{cm} \), \( u = -10 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-10} = \frac{1}{15} \Rightarrow \frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6} \).

\( v = 6 \, \text{cm} \) (virtual image).

5 cm
5.5 cm
6.5 cm
6 cm
4

A ray of light is incident from air (\( n = 1 \)) into glass (\( n = 1.5 \)) at \( 30^\circ \). What is the angle of refraction?

Snell’s law: \( n_1 \sin i = n_2 \sin r \).

Air (\( n_1 = 1 \)), glass (\( n_2 = 1.5 \)), \( i = 30^\circ \).

\( 1 \times \sin 30^\circ = 1.5 \times \sin r \).

\( \sin 30^\circ = 0.5 \Rightarrow 0.5 = 1.5 \sin r \Rightarrow \sin r = \frac{0.5}{1.5} = 0.333 \).

\( r = \sin^{-1}(0.333) \approx 19.5^\circ \).

19°
22°
25°
30°
1

A glass slab (\( n = 1.52 \)) of thickness \( 7.6 \, \text{cm} \) is placed over a dot. What is the apparent shift?

Shift = \( t \left( 1 - \frac{1}{n} \right) \).

\( t = 7.6 \, \text{cm} \), \( n = 1.52 \).

Shift = \( 7.6 \left( 1 - \frac{1}{1.52} \right) = 7.6 \left( 1 - 0.658 \right) = 7.6 \times 0.342 \approx 2.6 \, \text{cm} \).

2 cm
2.5 cm
3 cm
2.6 cm
4

A glass slab (\( n = 1.5 \)) of thickness \( 7.5 \, \text{cm} \) is placed over a point. What is the apparent shift?

Shift = \( t \left( 1 - \frac{1}{n} \right) \).

\( t = 7.5 \, \text{cm} \), \( n = 1.5 \).

Shift = \( 7.5 \left( 1 - \frac{1}{1.5} \right) = 7.5 \left( 1 - \frac{2}{3} \right) = 7.5 \times \frac{1}{3} = 2.5 \, \text{cm} \).

2 cm
2.2 cm
3 cm
2.5 cm
4

A fish is at a depth of \( 40 \, \text{cm} \) in water (\( n = 1.33 \)). What is its apparent depth when viewed from above?

Apparent depth = \( \frac{\text{real depth}}{n} \).

Real depth = \( 40 \, \text{cm} \), \( n = 1.33 \).

Apparent depth = \( \frac{40}{1.33} \approx 30.08 \, \text{cm} \).

30 cm
35 cm
40 cm
45 cm
1

An object is placed \( 24 \, \text{cm} \) from a concave mirror of focal length \( 12 \, \text{cm} \). What is the image distance?

Focal length: \( f = -12 \, \text{cm} \), \( u = -24 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-24} = \frac{1}{-12} \Rightarrow \frac{1}{v} = \frac{1}{-12} + \frac{1}{24} = \frac{-2 + 1}{24} = \frac{-1}{24} \).

\( v = -24 \, \text{cm} \) (real image).

18 cm
20 cm
24 cm
30 cm
3

A concave mirror of focal length \( 8 \, \text{cm} \) has an object placed \( 16 \, \text{cm} \) from it. What is the image distance?

Focal length: \( f = -8 \, \text{cm} \) (concave mirror).

Object distance: \( u = -16 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{v} + \frac{1}{-16} = \frac{1}{-8} \Rightarrow \frac{1}{v} = \frac{1}{-8} + \frac{1}{16} = \frac{-2 + 1}{16} = \frac{-1}{16} \).

\( v = -16 \, \text{cm} \) (real image).

16 cm
12 cm
8 cm
20 cm
1

A glass slab (\( n = 1.5 \)) of thickness \( 15 \, \text{cm} \) is placed over a pin. By how much does the pin appear raised?

Shift = \( t \left( 1 - \frac{1}{n} \right) \).

\( t = 15 \, \text{cm} \), \( n = 1.5 \).

Shift = \( 15 \left( 1 - \frac{1}{1.5} \right) = 15 \left( 1 - \frac{2}{3} \right) = 15 \times \frac{1}{3} = 5 \, \text{cm} \).

3 cm
5 cm
7 cm
10 cm
2

A lens has a power of \( -4 \, \text{D} \). What is its focal length?

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

\( P = -4 \, \text{D} \Rightarrow -4 = \frac{1}{f} \Rightarrow f = -\frac{1}{4} = -0.25 \, \text{m} = -25 \, \text{cm} \).

-20 cm
-22 cm
-25 cm
-30 cm
3

A convex lens of focal length \( 25 \, \text{cm} \) forms an image of an object placed \( 50 \, \text{cm} \) from it. What is the image distance?

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

Object distance: \( u = -50 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{v} - \frac{1}{-50} = \frac{1}{25} \Rightarrow \frac{1}{v} + \frac{1}{50} = \frac{1}{25} \).

\( \frac{1}{v} = \frac{1}{25} - \frac{1}{50} = \frac{2 - 1}{50} = \frac{1}{50} \).

\( v = 50 \, \text{cm} \).

25 cm
50 cm
75 cm
100 cm
2

In a concave mirror, what prevents the formation of an image when the object is placed at the focal point?

When the object is at the focal point of a concave mirror, the reflected rays become parallel and do not converge to a point. This results in the image being formed at infinity, meaning no distinct image forms at a finite distance.

Convergence of rays at the pole
Parallel reflected rays forming image at infinity
Divergence of rays behind the mirror
Absorption of light by the mirror
2

A telescope has an objective of focal length \( 180 \, \text{cm} \) and an eyepiece of focal length \( 6 \, \text{cm} \). What is its magnifying power?

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

\( f_o = 180 \, \text{cm} \), \( f_e = 6 \, \text{cm} \).

\( m = \frac{180}{6} = 30 \).

20
25
30
35
3

Why does a prism disperse white light into a spectrum of colors?

A prism disperses white light because different wavelengths (colors) of light have different refractive indices in the prism material. Shorter wavelengths (e.g., violet) refract more than longer wavelengths (e.g., red), causing the light to split into a spectrum as it exits the prism.

Due to total internal reflection
Due to equal refraction of all wavelengths
Due to different refractive indices for different wavelengths
Due to reflection at the prism surfaces
3

A convex mirror of focal length \( 18 \, \text{cm} \) produces an image \( 6 \, \text{cm} \) behind the mirror. What is the object distance?

Focal length: \( f = 18 \, \text{cm} \) (convex mirror).

Image distance: \( v = 6 \, \text{cm} \) (virtual image).

Mirror equation: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{6} + \frac{1}{u} = \frac{1}{18} \Rightarrow \frac{1}{u} = \frac{1}{18} - \frac{1}{6} = \frac{1 - 3}{18} = \frac{-2}{18} = \frac{-1}{9} \).

\( u = -9 \, \text{cm} \).

12 cm
9 cm
18 cm
6 cm
1

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