Ray Optics Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

Focal length: \( f = \frac{R}{2} = \frac{-20}{2} = -10 \, \text{cm} \) (concave mirror).

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

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

A ray of light passes from water (\( n = 1.33 \)) to air at an angle of incidence of \( 50^\circ \). What happens?

Critical angle: \( \sin i_c = \frac{n_2}{n_1} = \frac{1}{1.33} \approx 0.752 \Rightarrow i_c \approx 48.75^\circ \).

Since \( i = 50^\circ > i_c \), total internal reflection occurs.

Refraction at 60°
Refraction at 45°
Total internal reflection
No change
3

An object of height \( 3 \, \text{cm} \) is placed \( 20 \, \text{cm} \) from a concave mirror of focal length \( 15 \, \text{cm} \). What is the height of the image?

\( f = -15 \, \text{cm} \), \( u = -20 \, \text{cm} \).

\( \frac{1}{v} + \frac{1}{-20} = \frac{1}{-15} \Rightarrow \frac{1}{v} = \frac{1}{-15} + \frac{1}{20} = \frac{-4 + 3}{60} = \frac{-1}{60} \).

\( v = -60 \, \text{cm} \).

Magnification: \( m = -\frac{v}{u} = -\frac{-60}{-20} = -3 \).

Image height: \( h' = m \times h = -3 \times 3 = -9 \, \text{cm} \) (inverted).

Magnitude = \( 9 \, \text{cm} \).

6 cm
7 cm
9 cm
12 cm
3

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

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

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

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

6 cm
6.5 cm
7 cm
7.2 cm
4

A prism of angle \( 50^\circ \) and refractive index \( 1.4 \) produces what minimum deviation?

For a thin prism: \( D_m = (n - 1) A \).

\( n = 1.4 \), \( A = 50^\circ \).

\( D_m = (1.4 - 1) \times 50 = 0.4 \times 50 = 20^\circ \).

15°
18°
22°
20°
4

A compound microscope has an objective of focal length \( 1 \, \text{cm} \) and eyepiece of focal length \( 4 \, \text{cm} \) with a tube length of \( 14 \, \text{cm} \). What is the magnification at infinity?

Objective magnification: \( m_o = \frac{L}{f_o} = \frac{14}{1} = 14 \).

Eyepiece magnification: \( m_e = \frac{D}{f_e} = \frac{25}{4} = 6.25 \).

Total magnification: \( m = m_o \times m_e = 14 \times 6.25 = 87.5 \).

80
85
90
87.5
4

A light ray passes from water (\( n = 1.33 \)) to glass (\( n = 1.5 \)) at an angle of incidence of \( 30^\circ \). What is the angle of refraction?

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

\( 1.33 \sin 30^\circ = 1.5 \sin r \).

\( \sin 30^\circ = 0.5 \Rightarrow 1.33 \times 0.5 = 1.5 \sin r \Rightarrow 0.665 = 1.5 \sin r \).

\( \sin r = \frac{0.665}{1.5} \approx 0.443 \Rightarrow r = \sin^{-1}(0.443) \approx 26.3^\circ \).

20°
26°
30°
35°
2

What is the critical angle for a dense flint glass (\( n = 1.62 \)) to air interface?

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

Glass (\( n_1 = 1.62 \)), air (\( n_2 = 1 \)).

\( \sin i_c = \frac{1}{1.62} \approx 0.617 \).

\( i_c = \sin^{-1}(0.617) \approx 38.1^\circ \).

38°
42°
45°
50°
1

A concave lens of focal length \( 20 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) from the lens. What is the object distance?

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

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

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

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

\( u = -\frac{40}{3} \approx -13.33 \, \text{cm} \).

12 cm
13.3 cm
15 cm
20 cm
2

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

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

Image distance: \( v = 20 \, \text{cm} \) (real image).

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

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

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

15 cm
20 cm
25 cm
30 cm
2

A concave mirror has a radius of curvature of \( 40 \, \text{cm} \). An object of height \( 2 \, \text{cm} \) is placed \( 30 \, \text{cm} \) in front of it. What is the height of the image formed?

Focal length: \( f = \frac{R}{2} = \frac{-40}{2} = -20 \, \text{cm} \) (negative for concave mirror).

Object distance: \( u = -30 \, \text{cm} \).

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

\( \frac{1}{v} + \frac{1}{-30} = \frac{1}{-20} \Rightarrow \frac{1}{v} = \frac{1}{-20} + \frac{1}{30} = \frac{-3 + 2}{60} = \frac{-1}{60} \).

\( v = -60 \, \text{cm} \) (real image).

Magnification: \( m = -\frac{v}{u} = -\frac{-60}{-30} = -2 \).

Image height: \( h' = m \times h = -2 \times 2 = -4 \, \text{cm} \) (inverted, so negative).

Magnitude of height = \( 4 \, \text{cm} \).

2 cm
3 cm
4 cm
5 cm
3

An object is at a depth of \( 26.6 \, \text{cm} \) in a medium with refractive index \( 1.33 \). What is the apparent depth?

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

Real depth = \( 26.6 \, \text{cm} \), \( n = 1.33 \).

Apparent depth = \( \frac{26.6}{1.33} = 20 \, \text{cm} \).

20 cm
25 cm
30 cm
15 cm
1

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