Ray Optics Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A converging beam meets a convex lens (\( f = 10 \, \text{cm} \)) \( 5 \, \text{cm} \) before the convergence point. What is the new image distance?

Object distance: \( u = -5 \, \text{cm} \) (virtual object), \( f = 10 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{-5} = \frac{1}{10} \Rightarrow \frac{1}{v} + \frac{1}{5} = \frac{1}{10} \).

\( \frac{1}{v} = \frac{1}{10} - \frac{1}{5} = \frac{1 - 2}{10} = \frac{-1}{10} \).

\( v = -10 \, \text{cm} \) (10 cm to the left).

8 cm
9 cm
11 cm
10 cm
4

In a prism, when light is incident at a very small angle, what is the approximate relationship between deviation and the prism’s properties?

For a thin prism with a small angle of incidence, the deviation is approximately equal to the product of the prism’s refractive index minus one and the prism angle (D ≈ (n - 1)A). This simplification holds because the angles of refraction are small, minimizing higher-order effects.

Deviation ≈ (n - 1) × prism angle
Deviation ≈ n × prism angle
Deviation ≈ prism angle / (n - 1)
Deviation ≈ n / prism angle
1

A ray of light passes from air (\( n = 1 \)) to glass (\( n = 1.62 \)) at an angle of incidence of \( 30^\circ \). What is the angle of refraction?

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

Air (\( n_1 = 1 \)), glass (\( n_2 = 1.62 \)), \( i = 30^\circ \).

\( 1 \times \sin 30^\circ = 1.62 \times \sin r \).

\( \sin 30^\circ = 0.5 \Rightarrow 0.5 = 1.62 \sin r \Rightarrow \sin r = \frac{0.5}{1.62} \approx 0.309 \).

\( r = \sin^{-1}(0.309) \approx 18^\circ \).

18°
20°
25°
30°
1

An object of height \( 3 \, \text{cm} \) is placed \( 15 \, \text{cm} \) from a concave mirror of focal length \( 10 \, \text{cm} \). What is the height of the image?

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

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

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

Magnification: \( m = -\frac{v}{u} = -\frac{-30}{-15} = -2 \).

Image height: \( h' = m \times h = -2 \times 3 = -6 \, \text{cm} \) (inverted).

Magnitude = \( 6 \, \text{cm} \).

4 cm
5 cm
6 cm
8 cm
3

A convex lens (\( f = 40 \, \text{cm} \)) and a concave lens (\( f = 20 \, \text{cm} \)) are in contact. What is the effective focal length?

\( f_1 = 40 \, \text{cm} \), \( f_2 = -20 \, \text{cm} \).

\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{40} + \frac{1}{-20} = \frac{1 - 2}{40} = \frac{-1}{40} \).

\( f = -40 \, \text{cm} \) (diverging system).

-30 cm
-35 cm
-40 cm
-45 cm
3

An object is placed \( 16 \, \text{cm} \) from a convex mirror of focal length \( 24 \, \text{cm} \). What is the image distance?

Focal length: \( f = 24 \, \text{cm} \), \( u = -16 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-16} = \frac{1}{24} \Rightarrow \frac{1}{v} = \frac{1}{24} + \frac{1}{16} = \frac{2 + 3}{48} = \frac{5}{48} \).

\( v = \frac{48}{5} = 9.6 \, \text{cm} \) (virtual image).

8 cm
9 cm
10 cm
9.6 cm
4

In a concave mirror, when the object is placed at the center of curvature, where is the image formed?

For a concave mirror, when the object is at the center of curvature (C), the reflected rays converge back to the same point after reflection. This results in a real, inverted image formed at the center of curvature, with the same size as the object.

At the center of curvature
At the focal point
Between focal point and center of curvature
Behind the mirror
1

In a compound microscope, what role does the eyepiece play in the final image formation?

The eyepiece in a compound microscope acts as a magnifying lens, taking the real, inverted image formed by the objective and producing a larger, virtual image for the observer. It enhances the angular size of the intermediate image, making it appear magnified without altering its orientation.

Forms the initial real image
Inverts the image to make it erect
Reduces the image size
Magnifies the intermediate image
4

In a concave lens, what property of the lens determines the position of the virtual focal point?

The virtual focal point of a concave lens is where diverging rays appear to originate when traced backward. This position is determined by the lens’s focal length, which depends on its curvature and refractive index, defining the extent of divergence.

Lens thickness
Object distance
Focal length
Angle of incidence
3

A double convex lens of refractive index \( 1.55 \) has radii of curvature \( 22 \, \text{cm} \) and \( -22 \, \text{cm} \). What is its focal length?

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

\( n = 1.55 \), \( R_1 = 22 \, \text{cm} \), \( R_2 = -22 \, \text{cm} \).

\( \frac{1}{f} = (1.55 - 1) \left( \frac{1}{22} - \frac{1}{-22} \right) = 0.55 \left( \frac{1}{22} + \frac{1}{22} \right) = 0.55 \times \frac{2}{22} = \frac{1.1}{22} = \frac{1}{20} \).

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

18 cm
20 cm
25 cm
30 cm
2

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