Optics(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

OPTICS MCQs

Wave Optics, Ray Optics Refraction, Ray Optics Fundamentals, Ray Optics Curved Surface Refraction, Ray Optics Curved Mirrors

Total Questions : 111 | Page 4 of 12 pages

Question 31. In Young’s double slit experiment, the distance between the slits is 0.2 mm and the distance of the screen from the slits is 2 m. If light of wavelength 500 nm is used, find the number of fringes formed between two points separated by 1 cm on the screen.

Discuss Question

Answer: Option B. -> 2
:
B

y = n β

Here

β = \frac{λ D}{d} = 0.5 c m

2 fringes.

Question 32. The similarity between the sound waves and light waves is
[KCET 1994]

Both are electromagnetic waves
Both are longitudinal waves
Both have the same speed in a medium
They can produce interference

Discuss Question

Answer: Option D. -> They can produce interference
:
D
Sound wave and light waves both shows interference

Question 33. Of the two slits producing interference in Young’s experiment, one is covered with glass so that light intensity passing is reduced to 50%. Which of the following is correct?

Intensity of fringes remains unaltered
Intensity of bright fringe decreases and that of dark fringe increases
Intensity of bright fringe increases and that of dark fringe decreases
Intensity of both bright and dark fringes decreases

Discuss Question

Answer: Option B. -> Intensity of bright fringe decreases and that of dark fringe increases
:
B
Initial intensity of bright fringes =

(\sqrt{I} + \sqrt{I})^{2}

Final intensity of bright fringes after covering one slit with glass sheet =

{(\sqrt{I} + \sqrt{\frac{I}{2}})}^{2}

Clearly the intensity of bright fringes has decreased.
For dark fringes, initial intensity =

(\sqrt{I} - \sqrt{I})^{2} = 0

but new intensity

{(\sqrt{I} - \sqrt{\frac{I}{2}})}^{2} \neq 0

Clearly the intensity of dark fringes has increased.

Question 34. The distance between two slits is 1.2 mm. If the wavelength of light used is

4800 ˙ A

and the screen is at a distance of 1 m from the slits, how many fringes are observed on the screen in the space opposite to the slits

Discuss Question

Answer: Option C. -> 3
:
C

y = \frac{d}{2}; y = n β \Rightarrow \frac{d}{2} = n \frac{λ D}{d}

\Rightarrow n = 1.5

∴

the total fringes=3

Question 35. Two slits separated by 0.200 mm are to be used in Young’s double – slit experiment. Immediately behind the slits in a lens of focal length 0.500 m used to form the interference pattern on a screen located in the focal plane of the lens. What is wavelength of a monochromatic light source used to illuminate the slits if adjacent maxima of the interference pattern are separated by 1.00 mm?

400 nm
500 nm
600 nm
700 nm

Discuss Question

Answer: Option A. -> 400 nm
:
A

β = \frac{λ D}{d}, D = 0.5 m, d = 0.2 \times 10^{- 3} m, β = 1 \times 10^{- 3} m

So answer is 400 nm

Question 36. An electron beam is being used in YDSE. On increasing kinetic energy of the electrons, fringe width will:

Increase
Decrease
Remain unchanged
No interference pattern will be observed on the screen

Discuss Question

Answer: Option B. -> Decrease
:
B
Fringe width is directly proportional to wavelength of wave. If we increase the kinetic energy of theelectrons, since the de – Broglie wavelength of the electron beam will reduce, the fringe width will also decrease.

Question 37. In a Young's double - slit experience, the slits are 2mm apart and are illuminated with a mixture of two wavelengths

λ_{0} = 750 n m

and

λ = 900 n m

. The minimum distance from the common central bright fringe on a screen 2 m away from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other, is

1.5mm
3 mm
4.5 mm
6 mm

Discuss Question

Answer: Option C. -> 4.5 mm
:
C

y_{0} = n [\frac{D λ}{d}]

and

y_{n}^{'} = n^{'} (\frac{D λ^{'}}{d})

Equating

y_{n}

and

y_{n}^{'}

,we get

\frac{n}{n^{'}} = \frac{λ^{'}}{λ} = \frac{900}{750} = \frac{6}{5}

Hence, the first position at which overlapping occurs is

y_{6} = y_{6} = \frac{6 (2) (750 \times 10^{- 9})}{2 \times 10^{- 3}} = 4.5 m m

Question 38. In Young's double slit experiment, if one of the slit is closed fully, what will be visible on the screen?

A slit-shaped bright spot
A YDSE interference pattern
A diffraction pattern
Nothing will be visible

Discuss Question

Answer: Option C. -> A diffraction pattern
:
C
If one of theslits is closed then YDSE interference fringes are not formed on the screen, so you might be tempted to say 'A slit-shaped bright spot' will beall that you can see on the screen, but no, think about what you just learnt, a fringe pattern is observed due to diffraction from the single slit that is open.

Question 39. Monochromatic green light of wavelength 5×10^-7 m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2 m away is

0.25 mm
0.1 mm
1.0 mm
0.01 mm

Discuss Question

Answer: Option C. -> 1.0 mm
:
C

Question 40. A beam of light consisting of two wavelengths 6500 A and 5200 A is used to obtain interference fringes in Young’s double slit experiment. The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm. Find the distance of the