## Optics

## Light Wave And Interference Of Light Multiple Choice Question and Answers

**Question 1. In Young’s double-slit experiment, the source is white light If one of the slits is covered by a red filter and the other by a blue filter, then**

- There shall be alternate interference patterns of red and screen blue
- There shall be an interference pattern for red distinct from that for blue
- There shall be no interference fringes
- There shall be an interference pattern for red mixing

**Answer:** 3

**Question 2. A standard two-slit arrangement with slits S _{1}, S_{2}. P_{1} and P_{2} are two minima points on either side of P. At P_{2} on the screen, there is a hole and behind P_{2} is a second two-slit arrangement with slits S_{3}, S_{4} and a second screen B behind them.**

- There would be no interference pattern on B but it would be lighted ‘
- The screen B would be totally dark
- There would be a single bright point on the screen BC
- Thre would be a regular two-slit pattern on the screen B

**Answer:** 4

**Question 3. Two sources S _{1} and S_{2} of intensity I_{1} and I_{2} are placed in front of a screen. The pattern of intensity distribution seen in the central position is given in. Which of the following statements are true**

**Read and Learn More Class 12 Physics Multiple Choice Questions**

- S
_{1}and S_{2}have the same intensities - S
_{1}and S_{2}have a constant phase difference - S
_{1}and S_{2}have the same phase - S
_{1}and S_{2}have the same wavelength

**Answer:** 1, 2, and 3

**Question 4. For light diverging from a point source**

- The wavelength is spherical
- The intensity decreases in proportion to the distance squared
- The wavefront is parabolic
- The intensity of the wavefront does not depend on the distance

**Answer:** 1,2

**1. By Huygens’ wave theory of light, we cannot explain the phenomenon of**

- Interference
- Diffraction
- Photoelectric effect
- Polarisation

**Answer:** 3. Photoelectric effect

**2. By a monochromatic wave, we mean**

- A single ray
- A single ray of a single colour
- Wave having a single wavelength
- Many rays of a single colour

**Answer:** 3. Wave having a single wavelength

**3. If the distance between a point source and the screen is doubled, then the intensity of light on the screen will become**

- Four
- Double
- Half
- One fourth

**Answer:** 4. One fourth

**Question 5. Spherical wavefronts, emanating from a point source, strike a plane reflecting the source. What will happen to these wavefronts immediately after reflection?**

- They will remain spherical with the same curvature
- They will become plane wavefronts
- They will remain spherical with the same curvature, but a sign of curvature reversed
- They will remain spherical but with different curvature both in magnitude and sign

**Answer:** 3. They will remain spherical but with different curvature both in magnitude and sign

**Question 6. The maximum and minimum intensities of the resultant wave due to the superposition of two waves having intensities I and 4I will be**

- 5I and 3I
- 9I and I
- 9I and 3I
- 5I and I

**Answer: **2. 9I and I

**Question 7. If the ratio of maximum and minimum intensities of the interference fringes obtained in Young’s double slit experiment is 4: 1, then the ratio of the amplitudes of the two coherent sources will be**

- 4: 1
- 3:1
- 2:1
- 1:1

**Answer:** 2. 3:1

**Question 8. If the waves coming out from two sources of light having intensities I and 4I undergo interference, then the intensity at the points in the region of superposition where phase difference becomes \(\pi / 2\) is **

- I
- 3I
- 5I
- \(\frac{5 I}{2}\)

**Answer:** 3. 5I

**Question 9. Two monochromatic lights coming out from two coherent sources can produce constructive interference when their phase difference becomes**

- \(\frac{3 \pi}{2}\)
- 2pi
- pi
- \(\frac{\pi}{2}\)

**Question 10. Two waves of intensities **I_{1} and** **I_{2}**, cross a place in direction and same time. The summation of maximum and minimum intensities is**

- I
_{1}+I_{2} - \(\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)
- \(\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)
- 2(I
_{1}+I_{2})

**Answer:** 4. 2(I_{1}+I_{2})

**Question 11. The energy in the interference fringe**

- Produced in bright band
- Destroyed in the dark band
- Remainsto bright conserved, band only changes place from a dark band
- All the above

**Answer:** 3

**Question 12. The ratio of amplitudes of two waves, emitted from two coherent sources is 2: 1. If these two waves get superposed, the ratio of maximum and minimum intensity will be**

- 2
- 4
- 9
- 18

**Answer:** 3. 9

**Question 13. A double slit experiment where P and Q are the slits. The path lengths PX and QX are nλ and (n +2)λ A respectively. Taking the central fringe as zero, what is formed at XI (n is a whole number and A is wavelength)**

- First bright
- First dark
- Second Bright
- Second dark

**Answer:** 3. Second Bright

**Question 14. If two waves represented by y _{1}= 4 sin ωt and y_{2} = 3 sin\(\left(\omega t+\frac{\pi}{3}\right)\) interfere at a point, the amplitude of the resulting wave will about**

- 7
- 6
- 5
- 3.5

**Answer:** 2. 6

**Question 15. Maximum intensity in Young’s double-slit experiment is 70. If one slit is closed, the intensity would be**

- \(\frac{I_0}{4}\)
- \(\frac{I_0}{3}\)
- \(\frac{I_0}{2}\)

**Answer:** 2. \(\frac{I_0}{4}\)

**Question 16. If white light is used instead of monochromatic light in Young’s double slit experiment, then what change in the fringe width will be observed?**

- Interference fringes will disappear
- No change in interference fringes will occur
- Interference fringes become colourful
- The central line of the interference fringe will be of yellow colour

**Answer:** 3. Interference fringes become colourful

**Question 17. If the interference fringe width of the dark band is β _{1 }and that of the bright band is β_{2} then**

- 2β
_{1}= β_{2} - 2β
_{2}= β_{1} - β
_{1}= β_{2} - β
_{1}+3 β_{2 }= 1

**Answer:** 3. β_{1} = β_{2}

**Question 18. The distance between the first bright band and the first dark band in interference fringe [symbols have usual meaning ] is**

- \(\frac{\lambda D}{2 d}\)
- \(\frac{\lambda D}{d}\)
- \(\frac{\lambda D}{d}\)
- \(\frac{2 \lambda D}{d}\)

**Answer:** 3. \(\frac{\lambda D}{d}\)

**Question 19. In an ideal double slit experiment, when a glass plate, having a refractive index of 1.5 and thickness t, is placed in the path of interfered light rays of wavelength A then the intensity remains unchanged at the place where the cen¬ tral maximum was previously formed. Then the minimum thickness of the glass plate is**

- 2 λ
- \(\frac{2 \lambda}{3}\)
- \(\frac{\lambda}{3}\)
- λ

**Answer:** 1. 2 λ

**Question 20. In Young’s double-slit experiment, an electron ray is used. If the velocity of the electron is increased, the fringe width**

- Will Increase
- Will decrease
- Will remain the same
- Fringes will not be visible

**Answer:** 2. Will decrease

**Question 21. In Young’s double slit experiment, the fringe width is found to be 0.4mm. Now if the experiment is conducted by immersing the whole experiment set-up is water (p = 1.33), the magnitude of fringe width would be**

- 0.25mm
- 0.30mm
- 0.40mm
- 0.53 mm

**Answer:** 4. 0.53 mm

**Question 22. In Young’s double slit experiment, the fringe width is found to be 0.3 mm. Now a thin glass plate of refracting index 1.5 is placed in the path of any one of light rays coming from the slits, then the width of the fringe will be**

- 0
- 0.3 mm
- 0.45 mm
- 0.15 mm

**Answer:** 2. 0.3 mm

**Question 23. In Young’s double-slit experiment with monochromatic light, interference fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10 ^{-15} m towards the slit, the change in fringe width is 3× 10^{-5}m. If the distance between the slits is 10^{-3} m, the wavelength of the light used is**

- 6000 A°
- 5000 A°
- 3000 A°
- 4500 A°

**Answer:** 1. 6000 A°

**Question 24. In Young’s double slit experiment, the magnitude of the fringe width is β. If the whole set-up of the experiment is immersed in a liquid of refractive index p., then the magnitude of the fringe width will be**

- μ β
- \(\frac{\beta}{\mu+1}\)
- \(\frac{\beta}{\mu-1}\)
- \(\frac{\beta}{\mu}\)

**Answer:** \(\frac{\beta}{\mu}\)

**Question 25. In Young’s double Slit experiment, the distance between the slits is d. An Interference fringe is obtained on a screen placed at a distance D from the slits. A dark band is noticed just opposite to one of the slits. The wavelength of the light used is**

- \(\frac{D^2}{2 d}\)
- \(\frac{d^2}{2 D}\)
- \(\frac{D^2}{d}\)
- \(\frac{d^2}{D}\)

**Answer:** 4. \(\frac{d^2}{D}\)

**Question 26. Young’s double-slit experiment is conducted using green, red and blue lights in succession. The magnitude of fringe widths obtained are ****β _{G} , β_{R}, β_{B}, respectively. The observations will be**

**Answer:**

- β
_{G}> β_{B}> β_{R} - β
_{B}>β_{G}> β_{R} - β
_{R }>β_{B}>β_{G} - β
_{R }>β_{G}>β_{B}

**Answer: **4. β_{R }>β_{G} >β_{B}

**Question 27. What is not true for interference oflight?**

- The two sources must be coherent this is a very important condition
- The intensity of the dark band of the interference fringe may not be zero
- In interference, energy is not destroyed, only change of place takes place
- The intensity of the dark band of the interference fringe must be zero

**Answer:** 4

**Question 28. The fringe width of an interference fringe is y for the two given waves.If the frequency of the source becomes double then the fringe width would have been**

- \(\frac{1}{2}\) y
- y
- 2y
- \(\frac{3}{2}\) y

**Answer: **1. **Answer:**

**Question 29. For red and blue colours oflight, the distance of mth interfering fringe in Young’s double slit experiment from the central maxima be X _{mr} and X_{mb}. Correct relation is**

- X
_{mr}>X_{mb} - X
_{mr}>X_{mb} - X
_{mr}= X_{mb} - X
_{mr}+ X_{mb}= 0

**Answer:** 1. X_{mr}>X_{mb}

**Question 30. Two sources of wave S _{1} and S_{2}, having zero phase difference, coming from an isolated light source, produce interference. If the common wavelength of the two sources be A then it is observed that a destructive interference has taken place at point P. The value of (S_{2}P-S_{1}P) is**

- \(\frac{3 \lambda}{4}\)
- \(\frac{11 \lambda}{2}\)
- 2 λ
- 5 λ

**Answer: ** 2. \(\frac{11 \lambda}{2}\)

**Question 31. For the wavelength λ _{1} in Young’s double slit experiment, the seventh bright spot is situated at a distance d_{1} from the central bright spot. In the same experiment for the same number of bright spot and for the different wavelength λ_{2}, the distance between two bright spots is d_{2}. Now the value**

- \(\frac{\lambda_1}{\lambda_2}\)
- \(\frac{\lambda_2}{\lambda_1}\)
- \(\frac{\lambda_1^2}{\lambda_2^2}\)
- \(\frac{\lambda_2^2}{\lambda_1^2}\)

**Answer:** 1. \(\frac{\lambda_1}{\lambda_2}\)

**Question 32. In Young’s double-slit experiment, the distance between the two slits is d and the wavelength of the light used is A. The angular width of the fringe**

- \(\frac{d}{\lambda}\)
- \(\frac{\lambda}{d}\)
- \(\frac{2 \lambda}{d}\)
- \(\frac{\lambda}{2 d}\)

**Answer:** 4. \(\frac{\lambda}{2 d}\)

**Question 33. In Young’s double slit experiment, the two slits are d distance apart. The interference pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wave¬ length oflight is**

- \(\frac{D^2}{2 d}\)
- \(\frac{d^2}{2 D}\)
- \(\frac{D^2}{d}\)
- \(\frac{d^2}{D}\)

**Answer:** 4. \(\frac{d^2}{D}\)

**Question 34. Which of the following properties of light support the nature of light?**

- Light obeys laws reflection
- Tight shows interference
- Tight shows photoelectric effect
- The speed of light in water is smaller than that in

**Answer:** 2 And 4

**Question 35. Huygens’ principle of secondary wavelets may be used to**

- Find the specific of tight in the vacuum
- Explain the particle behaviour of light
- Find the subsequent position of a wavefront
- Explain Snell’s law

**Answer:** 2 And 4

**Question 36. When light travels from air to glass, a change occurs in its**

- Wavelength
- Frequency
- Speed
- Amplitude

**Answer:** 1, 3, and 4

**Question 37. White light is used in Young’s double-slit experiment. The separation between the slits is b and the screen is at a distance d (d >> b) from the die slits. At a point on the screen directly in front of the sllÿs, certain wavefronts are missing some of these wavelength are.**

- \(\lambda=\frac{b^2}{d}\)
- \(\lambda=\frac{2 b^2}{d}\)
- \(\lambda=\frac{b^2}{3 d}\)
- \(\lambda=\frac{2 b^2}{3 d}\)

**Answer:** 1, 3

**Question 38. If Young’s double-slit experiment Is performed using white light, then**

- The central fringe will be white
- No fringe will be completely dark
- The fringe adjacent to the central one will be red
- The fringe adjacent to the central node will be violet

**Answer:** 1,2, 4

**Question 39. In the young double slit experiment the light of wavelength λ _{2} fringe width is y_{1} and the light of wavelength λ_{2} fringe width is y_{2}. If the whole arrangement is dipped into a liquid of refractive index μ, it is found diat for the wavelength λ_{1}, fringe width becomes y_{3}. Now the correct relation is**

- \(y_2=y_1 \frac{\lambda_1}{\lambda_2}\)
- \(y_2=y_1 \frac{\lambda_2}{\lambda_1}\)
- \(y_3=\frac{y_1}{\mu}\)
- \(y_3=\mu y_1\)

**Answer:** 2, 3

**Question 40. Interference fringes may be observed due to superposition**

- (1) and (2)
- (2) and (4)
- (1) and (3)
- (3) and (4)

**Answer:** 1, 4

**Four light waves are represented by**

(1) y = a_{1} sinωt

(2) y = a_{2}sin(ωt + e)

(3) y = a_{2}sin 2ωt

(4) y = a_{2}sin2(ωt + e)

**Question 41. In Young’s double slit experiment the ratio of intensities of bright and dark fringes is 9. This means**

- The ratio of the intensities of individual sources is 5: 4
- The ratio of the intensities of individual sources is 4: 1
- The ratio of the amplitudes of the light waves is 3: 1
- The ratio of their amplitudes is

**Answer**: 2, 4

**Question 42. In Young’s double-slit experiment let A and B be the two slits. A thin plate of thickness t and refractive index p is placed in front of A. Let £ be die fringe width. The central maximum will shift**

- Towards A
- Towards B
- \(\text { by } t(\mu-1) \frac{\beta}{\lambda}\)
- \(\text { by } \mu t \frac{\beta}{\lambda}\)

**Answer:** 1,3

**Question 43. In Young’s double-slit experiment the distance of the screen from the plane of the slits is 1.0 m. The wavelength of light used and width of the fringe are 6000 A and 2 mm respectively.**

**1. Distance between the slits is**

- 0.5 mm
- 0.4 mm
- 0.2 mm
- 0.3 mm

**Answer: ** 4. 0.3 mm

**2. If the wavelength of the light used is 4800 A, the width of the fringe will be**

- 6 mm
- 2.2 mm
- 2.0 mm
- 1.8 mm

**Answer: **1. 6 mm

**Question 44. In Young’s double slit experiment the slits apart by 3 illuminated with a source of monochromatic light of wavelength 6000 A°. Interference fringes are obtained on a screen at a distance of 1 m from the slits.**

**1. The width of the fringe will be**

- 0.15 mm
- 0.2 mm
- 0.29 mm
- 0.12 mm

**Answer:** 2. 0.2 mm

**2. If the whole experimental arrangements are immersed in Aater the fringe width will, be**

- 0.15 mm
- 0.20 m
- 29 mm
- 0.12 mm

**Answer:** 1. 0.15mm

**Question 45. When waves from two coherent sources of amplitudes a and b superimpose , the amplitute R of the resultant wave is given by R, \(=\sqrt{a^2+b^2+2 a b \cos \phi}\)**

4. where <p is the constant phase angle between the two waves. The resultant intensity l is directly proportional to the square of the amplitude of the resultant wave i.e.,

I∝ (a^{2} + b^{2} + 2abcosΦ)

For constructive interference,

Φ= 2nπ. and I_{max} = (a + b)^{2}

For destructive interference,

Φ = – (2n – 1 )π and I_{max} = (a – b)^{2}

If I_{1}, I_{2} are intensities from two slits of widths ω_{1} and ω_{2} then

⇒ \(\frac{J_1}{I_2}=\frac{w_1}{w_2}=\frac{a^2}{b^2}\)

Light waves from two coherent sources of intensity ratio 81: 1 produce interference. With the help of the passage choose the most appropriate alternative for each of the following questions.

**1. The ratio of amplitudes of two sources is**

- 9:1
- 81:1
- 1:9
- 1: 81

**Answer: ** 1.9:1

**The ratio of slit widths of the two sources is**

- 9:1
- 81: 1
- 1:9
- 1: 81

**Answer:** 2. 81: 1

**3. The Die ratio of maxima and minima in the interference I pattern is**

- 9: 1
- 81:1
- 1:9
- 1:81

**Answer:** 3. 1:9

**4. If two slits in Young’s experiment have width ratio 1: 4, the ratio of maximum and minimum intensity in the interference pattern would be**

- 1: 4
- 1:16
- 9:1
- 9:16

**Answer:** 3. 9:1

**Question 46. Two coherent monochromatic beams of intensities and 41 respectively are superposed. The maximum and minimum intensities in the resulting pattern are**

- 5I and 3I
- 9I and 3I
- 4I and I
- 9I and I

**Answer: **4. 9I and I

**Question 47. A thin plastic sheet of a refractive index of 1.6 is used to cover one of the slits of a double-slit arrangement. The central point on the screen Is now occupied by what would have been the 7th bright fringe before the plastic was used.If the wavelength of light is 600 nm, what is the thickness (in μm) of the plastic**

- 7
- 4
- 8
- 6

**Answer:** 1. 7

Fringe width of the seven fringes = 7 × \(\times \frac{D}{2 d} \lambda\)

Again, displacement of the central bright fringe for the plastic sheet of thickness t = \(\frac{D}{2 d}(\mu-1) t\) t

According to the question

⇒ \(7 \frac{D}{2 d} \lambda=\frac{D}{2 d}(\mu-1) t\)

Here, = 1.6 : = 600 nm = 0.6 mA

t = \(\frac{7 \lambda}{\mu-1}=\frac{7 \times 0.6}{1.6-1}\)

= 7m

**Question 48. Two monochromatic coherent light beams A and B have intensities L and respectively.If these beams are superposed, the maximum and minimum intensities will be**

- \(\frac{9 L}{4}, \frac{L}{4}\)
- \(\frac{5 L}{4}, 0\)
- \(\frac{5 L}{2}, 0\)
- \(2 L, \frac{L}{2}\)

**Answer:** 1. \(\frac{9 L}{4}, \frac{L}{4}\)

I_{max}= \(\left(\sqrt{L}+\sqrt{\frac{L}{4}}\right)^2=\frac{9 L}{4}\)

I_{min }= \(\left(\sqrt{L}-\sqrt{\frac{L}{4}}\right)^2=\frac{L}{4}\)

**Question 49. If Young’s double slit experiment is done with white light, which of the following statements will be true?**

- All the bright fringes will be coloured
- All the bright fringes will be white
- The central fringe will be white
- No stable interference pattern will be visible

**Answer:** 3.

The central fringe will be white

If Young’s double slit experiment is performed with white light, the centre fringe will be white.

**Question 50. On a hot summer night, the refractive index of air Is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the ****Huygens’s principle leads us to conclude that as it travels, the light beam.**

- Becomeÿ narrower
- Goes horizontally without any deflection
- Bends downwards
- Bends upwards

**Answer:** 4. Bends upwards

Since air is comparatively lighter near the ground, the wavefronts do not remain parallel any more. Separation between two consecutive wavefronts is greater near the ground than at above. Since a ray of light is always normal to the wavefront, the light beam horizontal to the earth’s surface always bends upwards

**Question 51. In Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is**

- 1.56 mm
- 7.8 mm
- 9.75 mm
- 15.6 mm

**Answer:** 2. 7.8 mm

Let the m th bright fringe due to the wavelength 650 nm superposes with the n th bright fringe due to wavelength

Then

⇒ \(\frac{m \lambda_1 D}{d}=\frac{n \lambda_2 D}{d}\)

Or, \(\frac{m}{n}=\frac{\lambda_2}{\lambda_1}=\frac{520}{650}=\frac{4}{5}\)

In case the least distance

m = 4 and n= 5

The least distance = \(\frac{m \lambda D}{d}\)

= \(\frac{4 \times\left(650 \times 10^{-7}\right) \times 150}{0.5 \times 10^{-1}}\) cm

= 7.8 mm

**Question 52. In Young’s double-slit experiment, the Intensity of light on the screen whore the path difference μ Is k (λ being the wavelength of light used). The Intensity at a point where the path difference Is \(\frac{\lambda}{4}\) will be**

- k
- \(\frac{k}{4}\)
- \(\frac{k}{2}\)
- Zero

**Answer: **3. \(\frac{k}{2}\)

We know, I = \(l_0 \cos ^2 \frac{\pi \delta}{\lambda}\)

When the path difference is λ then, I = k = I_{0 }cos²π

When the path difference is \(\frac{\lambda}{4}\) , then I = I \(\cos ^2 \frac{\pi}{4}=\frac{I_0}{2}=\frac{k}{2}\)

**Question 53. The intensity at the maximum in Young’s double slit experiment is I _{0}. The separation between two slits is d = 5λ, where A is the wavelength oflight used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?**

- \(\frac{I_0}{4}\)
- \(\frac{3}{4} I_0\)
- \(\frac{I_0}{2}\)
- I
_{0}

**Answer:** 3. \(\frac{I_0}{2}\)

Path difference = \(\)

Φ = \(\frac{2 \pi}{\lambda} \times \text { path difference }\)

= \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\)

= \(\frac{\pi}{2}\)

Now,I =I \(I_0 \cos ^2 \frac{\phi}{2}\) = \(I_0 \cos ^2 \frac{\pi}{4}\)

= \(\frac{I_0}{2}\)

**Question 54. In Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light ****used is 5896A° and distance D between the screen and slits Is 100 cm. It is found that the angular width of the ****fringes is 0.20°. To increase the fringe angular width to 0.21° (with the same λ and D) the separation between the slits needs to be changed to**

- 2.1mm
- 1.9 mm
- 1.8mm
- 1.7mm

**Answer:** 2.1.9mm

Angular fringe width,

θ = \(\frac{\lambda}{d}\)

= \(\theta_1=\frac{\lambda}{d_1} \quad \text { or, } \theta_1 d_1=\lambda\) ………………………….. (1)

= \(\theta_2=\frac{\lambda}{d_2} \quad \text { or, } \theta_2 d_2=\lambda\) ………………………….. (2)

Dividing (2) by (1) we get,

\(\frac{\theta_2 d_2}{\theta_1 d_1}=\frac{\lambda}{\lambda}\) = 1

Or, \(d_2=\frac{\theta_1}{\theta_2} d_1=\frac{0.20}{0.21}\) × 2 = 1.9 mm