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MCQ Practice Test & Solutions: Chapter Test: Ray Optics and Optical Instruments - 2 (30 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Number of Questions: 30

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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 1


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 2


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 3


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 4


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 5

An object of height h = 5cm is located at a distance a = 12 cm from a concave mirror with focal length 10 cm. Find the height of the image.

Detailed Solution: Question 5


Chapter Test: Ray Optics and Optical Instruments - 2 - Question 6

An insect of negligible mass is sitting on a block of mass M, tied with a spring of force constant K. The block performs simple harmonic motion with amplitude A infront of a plane mirror placed as shown. The maximum speed of insect relative to its image will be


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 9

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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 10


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 11


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 12


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 13

A small candle is kept at a distance of 1.00 m from a screen. A lens of focal length 10.0 cm is placed between the candle and the screen. There are two positions of the lens that will result in a sharp image of the candle on the screen. The distance between these two positions is


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 14

Detailed Solution: Question 14

D is correct.

Let nair = 1 and nrod = 2/√3. By Snell's law at the flat end, sin r = sin θ / n,

where r is the angle of the refracted ray inside the rod measured from the axis.

For the ray to graze the inner curved wall the incidence at the curved surface must equal the critical angle.

The critical angle is θc = sin-1(1/n). For n = 2/√3, we have 1/n = √3/2, so θc = 60°.

In the meridional plane (plane containing the axis and the ray) the normal to the cylindrical surface is perpendicular to the axis, so the angle between the ray and the normal at the curved surface is 90° - r.

Setting this equal to the critical angle gives 90° - r = 60°, hence r = 30°.

Now use Snell's law:

sin r = sin θ / n

sin θ = n·sin r = (2/√3)·sin 30°

= (2/√3)·(1/2) = 1/√3.

Therefore θ = sin-1(1/√3), which is option D.

Chapter Test: Ray Optics and Optical Instruments - 2 - Question 15


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 16

Two coherent sources A & B emitting light of wavelength λ are placed at position (–D, 0) and (–D, 3λ) respectively. Here D >> λ. The number of minima on y-axis and maxima on xaxis respectively are


Detailed Solution: Question 16

Hence there are 6 minimas of y-axis and 5 maxima of x-axis.

Chapter Test: Ray Optics and Optical Instruments - 2 - Question 17

A ray of light strikes a plane mirror at an angle of incidence 45º as shown in the figure. After reflection, the ray passes through a prism of refractive index 1.5, whose apex angle is 4º. The angle through which the mirror should be rotated if the total deviation of the ray is to be 90º is


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 22

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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 24


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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 25

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Chapter Test: Ray Optics and Optical Instruments - 2 - Question 26

When light is reflected from a mirror a change occurs in its

Detailed Solution: Question 26

When a light wave is reflected from an object , it changes not only it's amplitude but also its phase according to the properties of the object at a particular point. 
Therefore, option A is the right answer.

Chapter Test: Ray Optics and Optical Instruments - 2 - Question 27

The images of clouds and trees in water always less bright than in reality

Detailed Solution: Question 27

The images of clouds and trees in water are always less bright than in reality because only a portion of the incident light is reflected and quite a large portion goes mid water.

Chapter Test: Ray Optics and Optical Instruments - 2 - Question 28

A rays is incident at an angle 38º on a mirror. The angle between normal and reflected ray is

Detailed Solution: Question 28

 As per the question, rays are falling on the mirror, which cleared that Regular reflection is taking place.
So, As per the laws of regular reflection.
=> ∠incident= ∠reflection
∠incident= ∠reflection=38Ao
Also we know that normal is at 90°to the mirror.
Now, to get the required angle we need to reduce angle of reflection from normal.
=>90Ao-38Ao=52Ao
so, the correct answer is option B.

Chapter Test: Ray Optics and Optical Instruments - 2 - Question 29

Mark the correct options

Detailed Solution: Question 29

If the final rays are converging, we have a real image.
This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
The correct answer is option B.

Chapter Test: Ray Optics and Optical Instruments - 2 - Question 30

Which of the following letters do not surface lateral inversion

Detailed Solution: Question 30

Lateral inversion-The phenomenon of left side appearing right side and right side appearing left side on reflection in a plane mirror is called lateral inversion.
Letters that do not show lateral inversion are,
A, H, I, M, O, T, U, V, W, X, Y
Hence, HOX will not show lateral inversion. 

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