Ray Optics Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

An object is placed \( 14 \, \text{cm} \) from a convex mirror of focal length \( 28 \, \text{cm} \). What is the magnification?

Focal length: \( f = 28 \, \text{cm} \), \( u = -14 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-14} = \frac{1}{28} \Rightarrow \frac{1}{v} = \frac{1}{28} + \frac{1}{14} = \frac{1 + 2}{28} = \frac{3}{28} \).

\( v = \frac{28}{3} \approx 9.33 \, \text{cm} \).

Magnification: \( m = -\frac{v}{u} = -\frac{9.33}{-14} \approx 0.666 \).

0.5
0.6
0.7
0.67
4

A ray of light passes from air (\( n = 1 \)) to water (\( n = 1.33 \)) at an angle of incidence of \( 35^\circ \). What is the angle of refraction?

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

Air (\( n_1 = 1 \)), water (\( n_2 = 1.33 \)), \( i = 35^\circ \).

\( 1 \times \sin 35^\circ = 1.33 \times \sin r \).

\( \sin 35^\circ \approx 0.574 \Rightarrow 0.574 = 1.33 \sin r \Rightarrow \sin r = \frac{0.574}{1.33} \approx 0.432 \).

\( r = \sin^{-1}(0.432) \approx 25.6^\circ \).

26°
30°
35°
40°
1

A concave lens has a focal length of \( 20 \, \text{cm} \). Where should an object be placed to get an image half the size of the object?

Focal length: \( f = -20 \, \text{cm} \) (negative for concave lens).

Magnification: \( m = \frac{h'}{h} = 0.5 \) (virtual, diminished).

\( m = \frac{v}{u} \Rightarrow 0.5 = \frac{v}{u} \Rightarrow v = 0.5u \).

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

\( \frac{1}{0.5u} - \frac{1}{u} = \frac{1}{-20} \Rightarrow \frac{2}{u} - \frac{1}{u} = -\frac{1}{20} \).

\( \frac{1}{u} = -\frac{1}{20} \Rightarrow u = -20 \, \text{cm} \).

10 cm
20 cm
30 cm
40 cm
2

What optical property of a prism allows it to be used in binoculars to invert images?

Prisms in binoculars use total internal reflection to invert and revert images. By reflecting light multiple times within the prism (e.g., in a Porro prism), the image orientation is corrected from the inverted form produced by the objective lens, maintaining the same size.

Refraction at small angles
Total internal reflection
Dispersion into colors
Convergence of light rays
2

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

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

What is the critical angle for a diamond-air interface if the refractive index of diamond is \( 2.42 \)?

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

Diamond (\( n_1 = 2.42 \)), air (\( n_2 = 1 \)).

\( \sin i_c = \frac{1}{2.42} \approx 0.413 \).

\( i_c = \sin^{-1}(0.413) \approx 24.4^\circ \).

20°
24°
28°
32°
2

A simple microscope uses a convex lens of focal length \( 5 \, \text{cm} \). What is the magnification when the image is at \( 25 \, \text{cm} \)?

Magnification: \( m = 1 + \frac{D}{f} \).

\( D = 25 \, \text{cm} \), \( f = 5 \, \text{cm} \).

\( m = 1 + \frac{25}{5} = 1 + 5 = 6 \).

4
5
6
7
3

An object is at a depth of \( 24 \, \text{cm} \) in a medium with refractive index \( 1.6 \). What is the apparent depth?

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

Real depth = \( 24 \, \text{cm} \), \( n = 1.6 \).

Apparent depth = \( \frac{24}{1.6} = 15 \, \text{cm} \).

15 cm
18 cm
20 cm
24 cm
1

A convex mirror of focal length \( 15 \, \text{cm} \) forms an image of an object placed \( 45 \, \text{cm} \) from it. What is the magnification of the image?

Focal length: \( f = 15 \, \text{cm} \) (positive for convex mirror).

Object distance: \( u = -45 \, \text{cm} \).

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

\( \frac{1}{v} + \frac{1}{-45} = \frac{1}{15} \Rightarrow \frac{1}{v} = \frac{1}{15} + \frac{1}{45} = \frac{3 + 1}{45} = \frac{4}{45} \).

\( v = \frac{45}{4} = 11.25 \, \text{cm} \) (virtual image).

Magnification: \( m = -\frac{v}{u} = -\frac{11.25}{-45} = 0.25 \).

0.2
0.25
0.33
0.5
2

An object is placed \( 15 \, \text{cm} \) in front of a concave mirror of focal length \( 10 \, \text{cm} \). Where is the image formed?

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

\( \frac{1}{v} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{-10} + \frac{1}{15} = \frac{-3 + 2}{30} = \frac{-1}{30} \).

\( v = -30 \, \text{cm} \) (real image).

20 cm
25 cm
30 cm
35 cm
3

What happens to light rays when they strike a convex mirror at an angle parallel to its principal axis?

In a convex mirror, rays parallel to the principal axis diverge after reflection. When traced backward, these diverging rays appear to originate from the focal point behind the mirror, which is why the focal point is virtual and located on the opposite side of the incident light.

They converge at the focal point
They reflect back along the same path
They appear to diverge from the focal point behind the mirror
They pass through the center of curvature
3

A concave lens of focal length \( 12 \, \text{cm} \) has an object placed \( 24 \, \text{cm} \) from it. What is the image distance?

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

Object distance: \( u = -24 \, \text{cm} \).

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

\( \frac{1}{v} - \frac{1}{-24} = \frac{1}{-12} \Rightarrow \frac{1}{v} + \frac{1}{24} = \frac{1}{-12} \Rightarrow \frac{1}{v} = \frac{1}{-12} - \frac{1}{24} = \frac{-2 - 1}{24} = \frac{-3}{24} = \frac{-1}{8} \).

\( v = -8 \, \text{cm} \) (virtual image).

6 cm
8 cm
10 cm
12 cm
2

A glass slab (\( n = 1.5 \)) of thickness \( 9 \, \text{cm} \) is placed over a mark. What is the apparent shift?

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

\( t = 9 \, \text{cm} \), \( n = 1.5 \).

Shift = \( 9 \left( 1 - \frac{1}{1.5} \right) = 9 \left( 1 - \frac{2}{3} \right) = 9 \times \frac{1}{3} = 3 \, \text{cm} \).

2 cm
2.5 cm
3.5 cm
3 cm
4

A convex mirror of focal length \( 10 \, \text{cm} \) forms an image \( 5 \, \text{cm} \) behind the mirror. What is the object distance?

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

Image distance: \( v = 5 \, \text{cm} \) (virtual image).

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

\( \frac{1}{5} + \frac{1}{u} = \frac{1}{10} \Rightarrow \frac{1}{u} = \frac{1}{10} - \frac{1}{5} = \frac{1 - 2}{10} = \frac{-1}{10} \).

\( u = -10 \, \text{cm} \).

5 cm
10 cm
15 cm
20 cm
1

A convex lens (\( f = 25 \, \text{cm} \)) and a concave lens (\( f = 50 \, \text{cm} \)) are in contact. What is the effective focal length?

\( f_1 = 25 \, \text{cm} \), \( f_2 = -50 \, \text{cm} \).

\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{25} + \frac{1}{-50} = \frac{2 - 1}{50} = \frac{1}{50} \).

\( f = 50 \, \text{cm} \) (converging system).

40 cm
45 cm
50 cm
55 cm
3

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