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  4. Ray Optics And Optical Instruments Chapter-Wise Test 2

Ray Optics And Optical Instruments Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A prism of angle \( 30^\circ \) and refractive index \( 1.6 \) produces what minimum deviation?

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

\( n = 1.6 \), \( A = 30^\circ \).

\( D_m = (1.6 - 1) \times 30 = 0.6 \times 30 = 18^\circ \).

15°
16°
20°
18°
4

What ensures that a convex mirror produces an image smaller than the object at all positions?

A convex mirror diverges reflected rays, making them appear to come from a point closer to the mirror than the object. This divergence reduces the image size relative to the object, resulting in a diminished image regardless of the object’s distance from the mirror.

Convergence of rays
Reflection at the focal point
Divergence reducing image size
Magnification exceeding unity
3

A simple microscope with a lens of focal length \( 10 \, \text{cm} \) forms an image at the least distance of distinct vision (\( 25 \, \text{cm} \)). What is the magnification?

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

\( D = 25 \, \text{cm} \), \( f = 10 \, \text{cm} \).

\( m = 1 + \frac{25}{10} = 1 + 2.5 = 3.5 \).

2.5
3.5
4.5
5.5
2

A compound microscope has an objective of focal length \( 1.25 \, \text{cm} \) and eyepiece of focal length \( 5 \, \text{cm} \) with a tube length of \( 15 \, \text{cm} \). What is the magnification at infinity?

Objective magnification: \( m_o = \frac{L}{f_o} = \frac{15}{1.25} = 12 \).

Eyepiece magnification: \( m_e = \frac{D}{f_e} = \frac{25}{5} = 5 \).

Total magnification: \( m = m_o \times m_e = 12 \times 5 = 60 \).

50
55
65
60
4

A convex lens of focal length \( 20 \, \text{cm} \) forms an image at \( 40 \, \text{cm} \) from the lens. What is the object distance?

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

Image distance: \( v = 40 \, \text{cm} \) (real image).

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

\( \frac{1}{40} - \frac{1}{u} = \frac{1}{20} \Rightarrow \frac{1}{u} = \frac{1}{40} - \frac{1}{20} = \frac{1 - 2}{40} = \frac{-1}{40} \).

\( u = -40 \, \text{cm} \).

30 cm
40 cm
50 cm
60 cm
2

A concave mirror of radius of curvature \( 30 \, \text{cm} \) has an object placed \( 45 \, \text{cm} \) from it. What is the image distance?

Focal length: \( f = \frac{R}{2} = \frac{-30}{2} = -15 \, \text{cm} \) (concave mirror).

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

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

\( \frac{1}{v} + \frac{1}{-45} = \frac{1}{-15} \Rightarrow \frac{1}{v} = \frac{1}{-15} + \frac{1}{45} = \frac{-3 + 1}{45} = \frac{-2}{45} \).

\( v = -\frac{45}{2} = -22.5 \, \text{cm} \) (real image).

22.5 cm
30 cm
15 cm
45 cm
1

A compound microscope has an objective of focal length \( 2 \, \text{cm} \) and tube length \( 20 \, \text{cm} \). If the eyepiece magnification is \( 5 \), what is the total magnification?

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

\( L = 20 \, \text{cm} \), \( f_o = 2 \, \text{cm} \Rightarrow m_o = \frac{20}{2} = 10 \).

Eyepiece magnification: \( m_e = 5 \) (given).

Total magnification: \( m = m_o \times m_e = 10 \times 5 = 50 \).

25
50
75
100
2

A prism of angle \( 40^\circ \) and refractive index \( 1.5 \) produces what minimum deviation?

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

\( n = 1.5 \), \( A = 40^\circ \).

\( D_m = (1.5 - 1) \times 40 = 0.5 \times 40 = 20^\circ \).

15°
18°
22°
20°
4

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

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

\( P = +3 \, \text{D} \Rightarrow 3 = \frac{1}{f} \Rightarrow f = \frac{1}{3} = 0.333 \, \text{m} \approx 33.33 \, \text{cm} \).

25 cm
30 cm
33 cm
40 cm
3

A converging beam meets a convex lens (\( f = 15 \, \text{cm} \)) \( 10 \, \text{cm} \) before the convergence point. What is the new image distance?

Object distance: \( u = -10 \, \text{cm} \) (virtual object), \( f = 15 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{-10} = \frac{1}{15} \Rightarrow \frac{1}{v} + \frac{1}{10} = \frac{1}{15} \).

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

\( v = -30 \, \text{cm} \) (30 cm to the left).

20 cm
25 cm
35 cm
30 cm
4

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