Wave Optics Test

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Wave Optics Test - 81

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Weekly Quiz Competition

Question 1
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A monochromatic beam of light falls on Young's double slit experiment apparatus as shown in figure. A thin sheet of glass is inserted .in front of lower slit $$S_{2}$$. If central bright fringe is obtained on screen at $$O$$.

A
$$(\mu - 1)t = d\sin \theta$$

B
$$(\mu - 1) t = d\cos \theta$$

C
$$(\mu - 1)t + d\sin \theta = 0$$

D
$$\dfrac {t}{\mu - 1} = \dfrac {d}{\sin \theta}$$
Question 2
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Directions For Questions

The YDSE apparatus is modified by placing an isotropic transparent plate of high melting point in front of one of the slits. The refractive index of the plate is $$\mu_{r} =1.5$$ at room temperature and its thickness is $$t = 2\mu m$$. The refractive index of plate will increase when temperature increases and temperature coefficient of refractive index of the plate (i.e., the fractional change in refractive index per unit rise in temperature) is $$2\times 10^{-6}/ ^{\circ}C$$. The incident light is having wavelength $$\lambda = 600\overset {\circ}{A}$$. The separation between the slits is $$d = 0.2\ cm$$, and separation between the slit and the screen is $$2\ m$$. Assume that slits are of equal intensity.
Based on the above information, answer the following questions:

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If the plate is heated so that it temperature rises by $$10^{\circ}C$$, then how many fringes will cross a particular point on the screen? (Neglect the thermal expansion of plate):

A
$$1000$$

B
$$10^{-4}$$

C
$$5000$$

D
$$\dfrac {1}{3000}$$

Solution
Question 3
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One of the two slits in YDSE is painted over, so that it transmits only light waves having intensity half of the intensity of the light waves through the other slit. As a result of this

A
fringe pattern disappears

B
bright fringes become brighten and dark one becomes darker

C
dark and bright fringes get fainter

D
dark fringes get brighter and bright fringes get darker

Solution

We know that,
$$I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}cos\triangle\phi$$

Here $$I_{1}=I_{0}$$
$$I_{2}=I_{0}/2$$

For maximum intensity,$$cos \triangle \phi=1$$

For minimum intensity, $$cos \triangle \phi=-1$$

Before painting,

$$I=I_{2}=I_{0}, I_{max}=4I_{0}, I_{min}=0$$

After painting,

$$I_{1}=I_{0},I_{max}=\left(\dfrac{3+2\sqrt{2}}{2}\right)I_{0} < 4I_{0}$$

$$I_{2}=\dfrac{I_{0}}{2}, I_{min}= \left(\dfrac{3-2\sqrt{2}}{2}\right)I_{0} > 0$$

Hence, dark fringes get brighter and bright fringes get darker.
Question 4
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The maximum intensity in Young's double-slit experiment is $$I_{0}$$. Distance between the slits is $$d=5\lambda$$, where $$\lambda$$ is the wavelength of monochromatic light is used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $$D=d$$ ?

A
$$\dfrac{I_{0}}{2}$$

B
$$\dfrac{3}{4}I_{0}$$

C
$$I_{0}$$

D
$$\dfrac{I_{0}}{4}$$

Solution

Path difference, $$\triangle x= \dfrac{yd}{D}$$

Here, $$y = \dfrac {5\lambda}{2}$$

and $$D= 10d = 50 \lambda$$ $$(as \quad d \quad = \quad 5\lambda)$$
So,

$$\triangle x= \left(\dfrac{5\lambda}{2}\right)\left(\dfrac{5\lambda}{50\lambda}\right)= \dfrac{\lambda}{4} $$

Corresponding phase difference will be
$$\phi= \left(\dfrac{2\pi}{\lambda}\right)(\triangle x)= \left(\dfrac{2\pi}{\pi}\right)\left(\dfrac{\lambda}{4}\right)= \dfrac{\pi}{2}$$

or $$\dfrac{\phi}{2}=\dfrac{\pi}{4}$$

$$ \therefore I= I_{0}cos^{2}\left(\dfrac{\phi}{2}\right)$$

$$=I_{0}cos^{2}\left(\dfrac{\pi}{4}\right)= \dfrac{I_{0}}{2}$$
Question 5
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Two coherent light sources, each of wavelength $$\lambda$$, are separated by a distance 3 $$\lambda$$. The maximum number of minima formed on line $$AB$$, which runs from $$-\infty$$ to $$+\infty$$, is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

Given,
d = 3 $$\lambda$$
Screen length varies from $$-\infty$$ to $$+\infty$$
Now we know that ,
For maxima $$\triangle x_{minimum}$$ = 0 and $$\triangle x_{maximun}$$ = d ( slit distance) = 3 $$\lambda$$

For nth minima $$ \triangle x$$ = $$\dfrac{(2n-1) \lambda}{2}$$ and minima must lies in between the maxima ranges i.e, {0 to 3 $$\lambda$$}.

$$\therefore$$ $$ \triangle x$$ = $$\dfrac {\lambda}{2}$$ n = 1

$$ \triangle x$$ = $$\dfrac {3\lambda}{2}$$ n = 2

$$ \triangle x$$ = $$\dfrac {5\lambda}{2}$$ n = 3

There can be three minima from central point to $$\infty$$ corresponding to $$\dfrac{\lambda}{3}, \dfrac{3\lambda}{2}, \dfrac{5\lambda}{2}$$ path differences.

Similarly, there can be three minima from central point to $$-\infty$$ corresponding to $$\dfrac{-\lambda}{3}, \dfrac{-3\lambda}{2}, \dfrac{-5\lambda}{2}$$ path differences.

Therefore, total number of minima= $$2n_{max}=6$$
Question 6
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For the situation shown in the figure below. $$BP - AP = \dfrac {\lambda}{3}$$ and $$D > > d$$. The slits are of equal widths, having intensity $$I_{0}$$. The intensity at $$P$$ would be :

A
$$4I_{0}$$

B
$$2I_{0}$$

C
$$3I_{0}$$

D
$$\dfrac {7}{2}I_{0}$$

Solution
Question 7
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The path difference between two interfering waves at a point on the screen is $$\lambda/6$$. The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit in same)

A
$$0.853$$

B
$$8.53$$

C
$$0.75$$

D
$$7.5$$

Solution

We know that at path difference = $$\dfrac{\lambda}{6}$$, phase difference = $$\dfrac{\pi}{3}$$

and, $$I=I_{0}+I_{0}+2I_{0}cos\dfrac{\lambda}{3}= 3I_{0}$$

So, the required ratio is $$\dfrac{3I_{0}}{4I_{0}}=0.75$$
Question 8
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In YDSE, if a bichromatic light having wavelengths $$\lambda_{1}$$ and $$\lambda_{2}$$ is used, then the maxima due to both lights will overlap at a certain distance $$y$$ of from the central maxima. Take separation between slits as $$d$$ and distance between screen and slit as $$D$$. Then, the value $$y$$ will be

A
$$\left({\lambda_{1}+\lambda_{2}}{2D}\right)d$$

B
$$\dfrac{\lambda_{1}-\lambda_{2}}{2D} \times 2d$$

C
LCM of $$\dfrac{\lambda_{1}D}{d}$$ and $$\dfrac{\lambda_{2}D}{d}$$

D
HCF of $$\dfrac{\lambda_{1}D}{d}$$ and $$\dfrac{\lambda_{2}D}{d}$$

Solution

We know that the distance of nth maxima from the central maxima is given by:-
$$y = \dfrac{n \lambda D}{d}$$

Let $$n_{1}^{th}$$ maxima corresponding to $$\lambda_{1}$$ wavelength be overlapping with $$n_{2}^{th}$$ maxima corresponding to $$\lambda_{2}$$ wavelength. Then the required distance,

$$y=\dfrac{n_{1}\lambda_{1}D}{d}=\dfrac{n_{2}\lambda_{2}D}{d}$$
Here, $$n_1$$ and $$n_2$$ are changing while $$\dfrac{ \lambda_1 D }{d} $$ and $$\dfrac{ \lambda_2 D }{d} $$ are constant.

Hence,
$$y$$=LCM of $$\dfrac{\lambda_{1}D}{d}$$ and $$\dfrac{\lambda_{2}D}{d}$$
Question 9
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When the object is self-luminous, the resolving power of a microscope is given by the expression

A
$$\dfrac{2\mu \sin \theta }{1.22\lambda}$$

B
$$\dfrac{\mu \sin \theta }{1\lambda}$$

C
$$\dfrac{2\mu \cos \theta }{1.22}$$

D
$$\dfrac{2\mu }{\lambda}$$

Solution
Question 10
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In Young's double slit experiment the source $$S$$ and the two slits $$A$$ and $$B$$ are vertical with slit $$A$$ above slit $$B$$. The fringes are observed on a vertical screen $$K$$. The optical path length from $$S$$ to $$B$$ is increased very slightly ( by introducing a transparent material of higher refractive index) and the optical path length from $$S$$ to $$A$$ is not changed, as a result the fringe system on $$K$$ moves

A
Vertically downward slightly

B
Vertically upwards slightly

C
Horizontally, slightly to the left

D
Horizontally, slightly to the right

Solution