Wave Optics Test

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Wave Optics Test - 67

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Question 1
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An astronomical telescope, consists of two thin lenses set $$36 cm$$ a part and has a magnifying power $$8$$. Calculate the focal length of the lenses.

A
$$32 cm$$

B
$$18 cm$$

C
$$25 cm$$

D
$$36 cm$$

Solution

Here, $$f_o+f_e=36 cm$$
$$M=-8$$ (magnifying power is negative)
Now, $$M=\cfrac {f_o}{f_e}$$
$$\therefore -8=-8=-\cfrac {f_o}{f_e}$$
$$\Rightarrow f_o=8f_e$$

From the equations, we have
$$8f_e+f_e=36$$ or $$f_e=4\ cm$$

Again, $$f_o=8f_e=8\times 4=32\ cm$$
Question 2
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Directions For Questions

The resolving power of a microscope is its capacity to form distinct images of two very close objects. This power depends upon the wavelength of the light used for illuminating the object and is more for violet light than for the red.
The breakthrough in high-resolution microscopy was made with the advent of quantum mechanics which established the dual nature of wave and matter. Ordinary light, under suitable conditions, could behave like a stream of particles, whereas, with the help of particles moving with high speeds all wave phenomena like diffraction, interference, etc. Could be demonstrated. The relationship between the momentum of the particle and the wavelength of the associated wave is given by the de Broglie's hypothesis, which states that the wavelength is inversely proportional to momentum.

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A source of electrons or neutrons can be located at $$S$$. The beam is defined by slits $$S_{1}$$ and $$S_{2}$$. The particles pass through two plates $$P_{1}$$ and $$P_{2}$$ such that the first plate is at zero potential and the second plate can be given any high potential. For using matter waves in a microscope of high resolution which of the following combination must be chosen?

A
Electron beam with high accelerating potential

B
Electron beam with low accelerating potential

C
Neutron beam with high accelerating potential

D
Neutron beam with low accelerating potential

Solution

Resolution is directly proportional to wavelength, but wavelength is inversely proportional to momentum(i.e., mass).
Electron has less mass than neutron, therefore we electron beam must be chosen.
As wave phenomenon is exhibited at high speed, therefore high accelerating potential has to be applied.
Question 3
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Two thin films of the same material but different thickness are separated by air. Monochromatic light in incident on the first film. When viewed normally from the point $$A$$, the second film appears dark.
From point $$B$$, on normal viewing

A
the first film will appear bright

B
the first film will appear dark

C
the second film will appear bright

D
the second film will appear dark

Solution

Due to Destructive Interference of light passing from Film 1 and falling on Film 2, The $$2^{nd}$$ film will appear dark from both sides.

Answer is D
Question 4
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$${M}_{1}$$ and $${M}_{2}$$ are plane mirrors and kept parallel to each other. At point O, there will be a maxima for wavelength $$\lambda$$. Light from a monochromatic source $$S$$ of wavelength $$\lambda$$ is not reaching directly on the screen. Then, $$\lambda$$ is:

A
$$\cfrac { { 3d }^{ 2 } }{ D } $$

B
$$\cfrac { { 3d }^{ 2 } }{ 2D } $$

C
$$\cfrac { { d }^{ 2 } }{ D } $$

D
$$\cfrac { { 2d }^{ 2 } }{ D } $$

Solution

Light emerging from point S gets reflected from mirrors $$M_{1}$$ and $$M_{2}$$, to interfere constructively at point O.
This means that path difference between light travelling $$SM_{1}O$$ and $$SM_{2}O$$ must be $$\lambda$$.
As shown in the figure, path $$SM_{1}O = 2\sqrt{(\dfrac{D}{2})^2 +(\dfrac{d}{2})^2}$$
path $$SM_{2}O = 2\sqrt{(\dfrac{D}{2})^2 + {d}^2}$$
$$2\sqrt{(\dfrac{D}{2})^2 + {d}^2}$$ $$-$$ $$2\sqrt{(\dfrac{D}{2})^2 +(\dfrac{d}{2})^2} = \lambda$$
Thus using Bionomial expansion, $$\lambda$$ comes out to be $$\dfrac{3d^2}{2D}$$.
Question 5
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A parallel beam of white light is incident on a thin film of air of uniform thickness. Wavelengths $$7200\mathring { A } $$ and $$5400\mathring { A } $$ are observed to be missing from the spectrum of reflected light viewed normally. The other wavelength in the visible region missing in the reflected spectrum is

A
$$6000\mathring { A } $$

B
$$4320\mathring { A } $$

C
$$5500\mathring { A } $$

D
$$6500\mathring { A } $$

Solution

The wavelength missing from the reflected spectrum must satisfy the condition
$$2\mu t=n\lambda $$
Where t is thickness of air film
When two wavelengths $${ \lambda }_{ 1 }$$ & $${ \lambda }_{ 2 }$$ are missing
$$2\mu t=n{ \lambda }_{ 1 }=\left( n+1 \right) { \lambda }_{ 2 }$$
or $$n\times 7200=\left( n+1 \right) 5400$$
$$n=3$$
The next missing wavelength must satisfy the condition
$$n{ \lambda }_{ 1 }=\left( n+2 \right) { \lambda }_{ 2 }$$
$$7200\times 3=\left( 3+2 \right) { \lambda }_{ 2 }$$
$$7200\times 3=5{ \lambda }_{ 2 }$$
$$\therefore { \lambda }_{ 2 }=4320A°$$
Question 6
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Two identical coherent sources are placed on a diameter of a circle of radius $$R$$ at separation $$x(<<R)$$ symmetrical about the center of the circle. The sources emit identical wavelength $$\lambda$$ each. The number of points on the circle of maximum intensity is $$(x=5\lambda$$):

A
$$20$$

B
$$22$$

C
$$24$$

D
$$26$$

Solution

For maxima: $$ d\sin { \theta } =n\lambda $$
On circle, for $$ -\dfrac { \pi }{ 2 } \le \theta \le \dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } +1$$
for $$ \dfrac { \pi }{ 2 } <\theta <-\dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } -1$$
Total number of maxima $$ n=\dfrac { 4d }{ \lambda } =\dfrac { 4\left( 5\lambda \right) }{ \lambda } =20 $$
Question 7
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Two waves of light in air have the same wavelength and are initially in phase. They then travel through plastic layer with thicknesses of $${L}_{1}=3.5mm$$ and $${L}_{2}=5.0mm$$ and indices of refraction $${n}_{1}=1.7$$ and $${n}_{2}=1.25$$ as shown in figure. The rays later arrive at a common point. The longest wavelength of light for which constructive interference occurs at the point is

A
$$0.8 mm$$

B
$$1.2 mm$$

C
$$1.7 mm$$

D
$$2.9 mm$$

Solution

Optical path travelled by wave 1, $$x_1=n_1L_1 +n_3L_3$$
$$L_3=1.5 mm , n_3=1$$
Optical path travelled by wave 2, $$x_2= n_2L_2$$
For wave 1 and wave 2 to interfere constructively, path difference $$ x_1 -x_2=m\lambda$$
For longest wavelength $$\lambda_L$$, $$m=1$$
$$(1.7\times3.5+1\times1.5)-1.25\times5=\lambda_L$$
Thus $$\lambda_L=1.2 mm$$
Question 8
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Two transparent slabs have the same thickness as shown in figure. One is made pf material $$A$$ of refractive index $$1.5$$. The other is made of two materials $$B$$ and $$C$$ with thickness in the ratio $$1:2$$. The refractive index of $$C$$ is $$1.6$$. If a monochromatic parallel beam passing through the slabs has the same number of wavelengths inside both, the refractive index of $$B$$ is

A
$$1.1$$

B
$$1.2$$

C
$$1.3$$

D
$$1.4$$

Solution

Given: $$n_A = 1.5$$; $$n_C = 1.6$$
Optical path length traveled by beam is same in both the slabs.
$$\therefore$$ $$n_At = n_B \times \dfrac{t}{3}+ n_C\times \dfrac{2t}{3}$$
$$\therefore$$ $$(1.5)t = n_B \times \dfrac{t}{3}+ (1.6)\times \dfrac{2t}{3}$$
$$\implies n_B = 1.3$$
Question 9
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Consider the optical system shown in the figure that follows. The point source of light $$S$$ is having wavelength equal to $$\lambda$$. The light is reaching screen only after reflection. For point $$P$$ to be $$2^{nd}$$ maxima, the value of $$\lambda$$ would be $$(D > > d$$ and $$d > > \lambda)$$ :

A
$$\dfrac {12d^{2}}{D}$$

B
$$\dfrac {6d^{2}}{D}$$

C
$$\dfrac {3d^{2}}{D}$$

D
$$\dfrac {24d^{2}}{D}$$

Solution

Two rays from light source S get reflected from the two mirrors at M and N respectively and then interfere at point P.
Length of path $$SMP= \sqrt{d^2 + (\dfrac{D}{2})^2} +\dfrac{D}{2}$$
Length of path $$SNP= \sqrt{(3d)^2 + (\dfrac{D}{2})^2} +\sqrt{(\dfrac{D}{2})^2+(4d)^2}$$
For second maxima to be obtained at P,the path difference must be $$2\lambda$$
i-e $$SNP-SMP= $$ $$ [\sqrt{(3d)^2 + (\dfrac{D}{2})^2} +\sqrt{(\dfrac{D}{2})^2+(4d)^2}]-$$ $$[ \sqrt{d^2 + (\dfrac{D}{2})^2} +\dfrac{D}{2}] = 2\lambda$$
Given d<<D
using Bionomial Expansion, $$\sqrt{1+x}=1+\dfrac{x}{2} for \,x<<1$$
$$\lambda=\dfrac{12d^2}{D}$$
Question 10
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Two points sources separated by $$2.0m$$ are radiating in phase with $$\lambda=0.50m$$. A detector moves in a circular path around the two sources in a plane containing them. How many maxima are detected?

A
$$16$$

B
$$20$$

C
$$24$$

D
$$32$$

Solution

For maxima $$ d\sin { \theta } =n\lambda $$
On circle, for $$ -\dfrac { \pi }{ 2 } \le \theta \le \dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } +1$$
for $$ \dfrac { \pi }{ 2 } <\theta <-\dfrac { \pi }{ 2 } \Rightarrow n=\dfrac { 2d }{ \lambda } -1$$
Total number of maxima $$ n=\dfrac { 4d }{ \lambda } =\dfrac { 4\left( 2 \right) }{ 0.5 } =16 $$