Wave Optics Test

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Wave Optics Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

Wavelength of light used in an optical instrument are $$\lambda_1=4000\overset{o}{A}$$ and $$\lambda_2=5000\overset{o}{A}$$, then ratio of their respective resolving powers(corresponding to $$\lambda_1$$ and $$\lambda_2$$) is?

A
$$16.25$$

B
$$9:1$$

C
$$4:5$$

D
$$5:4$$

Solution

Resolving Power (R.P) :It is the reciprocal of resolving limit<br>
$$R. P. = \dfrac{1} {d\theta}$$
For a Telescope : Resolving limit,
$$d\theta$$ $$=$$ $$\dfrac{1.22 \times \lambda}{a}$$
$$\therefore R. P. \propto \frac{1} {\lambda}$$
Where $$\lambda$$ = wavelength of light used to illuminate the object.
$$a$$ =aperture of the objective.
Given: $$\lambda_1 = 4000 A$$ and $$\lambda_2 = 5000A$$
$$R. P.| \propto \dfrac{1}{\lambda}| \Rightarrow \dfrac{(R.P.)_1}{(R.P.)_2}=\dfrac{\lambda_2}{\lambda_1}=\dfrac 54$$
Question 2
1 / -0

The sensor is exposed for $$0.1 s$$ to a $$200 W$$ lamp $$10 m$$ away. The sensor has opening that is $$20 mm$$ in a diameter. How many photons enter the sensor if the wavelength of light is $$600 mm$$? (assume that all the energy of the lamp is given off as light).

A
$$1.53 \times 10^{11}$$

B
$$1.53 \times 10^{2}$$

C
$$1.53 \times 10^{4}$$

D
$$1.53 \times 10^{13}$$

Solution

(Refer to Image)
Energy incident on sensor= $$n \cfrac {hc}{x}$$
$$=\cfrac {\pi\left(\cfrac {20}{1000}\right)\times \cfrac {1}{4}}{4 \pi (10)^2}\times 200 \times \cfrac {1}{10}=x \times \cfrac {6.626}{10^{34}}\times \cfrac {3 \times 100}{600}$$
$$\Rightarrow n= 1.53 \times 10^{13}$$
Question 3
1 / -0

An analyser is inclined to a polarizer at an angle of $$30^o$$. the intensity of light emerging from the analyser is $$\dfrac{1}{n}^{th}$$ of that is incident on the polarizer. Then n is equal to

A
$$4$$

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

C
$$\dfrac{8}{3}$$

D
$$\dfrac{1}{4}$$

Solution

Given,
$$\theta=30^0$$
Lets consider $$I_0=$$ Intensity of incident light
Intensity of light passing though the analyser is given by
$$I_a=\dfrac{I_0}{n}$$. . . . . .(1)
By malus law,
$$I'=Icos^2 \theta$$
$$\dfrac{I_0}{n}=\dfrac{I_0}{2}cos^2 30^0$$
$$\dfrac{1}{n}=\dfrac{1}{2}\times (\dfrac{\sqrt{3}}{2})^2=\dfrac{3}{8}$$
$$n=\dfrac{8}{3}$$
The correct option is C.
Question 4
1 / -0

Ratio of intensity of two waves is $$25 : 1$$. If interference occurs, then ratio of maximum and minimum intensity should be:-

A
$$25 : 1$$

B
$$5 : 1$$

C
$$9 : 4$$

D
$$4 : 9$$
Question 5
1 / -0

The wavefront of a lightbeam is given by the equation $$x+2y+3z=c$$,(where c is arbitary constant) the angle made by the direction of light with the y-axis is:

A
$${ cos }^{ -1 }\dfrac { 1 }{ \sqrt { 14 } } $$

B
$${ cos }^{ -1 }\dfrac { 2 }{ \sqrt { 14 } } $$

C
$${ sin }^{ -1 }\dfrac { 1 }{ \sqrt { 14 } } $$

D
$${ sin }^{ -1 }\dfrac { 2 }{ \sqrt { 14 } } $$

Solution

Given,
Direction of light, $$\vec n=\hat i+2\hat j+3\hat k$$
The angle made by the direction of light with y-axis is
$$cos\beta=\hat n.\hat j$$
$$cos\beta=\dfrac{\vec n}{|\vec n|}.\hat j$$
$$cos\beta=\dfrac{(\hat i+2\hat j+3\hat k).(0\hat i+\hat j+0\hat k)}{\sqrt{(1)^2+(2)^2+(3)^2}}=\dfrac{2}{\sqrt{14}}$$
$$\beta=cos^{-1}\dfrac{2}{\sqrt{14}}$$
The correct option is B.
Question 6
1 / -0

In Young's double slit experiment, the two slits are $$0.2 mm$$ apart. The interference fringes for light of wavelength $$6000 \mathring{A}$$ are found on the screen $$80 cm$$ away. The distance of fifth dark fringe, from the central fringe, will be:

A
$$6.8 \ mm$$

B
$$7.8 \ mm$$

C
$$9.8 \ mm$$

D
$$10.8 \ mm$$
Question 7
1 / -0

Two particles of masses $$m$$ and $$2m$$ are kept at a separation $$'d'$$. Then the distance of the point on the line joining the particle where the intensity is zero is:-

A
$$(\sqrt{2} + 1)d$$ from $$'2m'$$

B
$$(\sqrt{2} - 1)d$$ from $$'2m'$$

C
$$\dfrac{3d}{4}$$ from $$'m'$$

D
$$(\sqrt{2} - 1)d$$ from $$'m'$$

Solution

Refer above image,
$$M_0$$ is the mass of earth
$$m$$ is the mass of one particle
$$2m$$ is the mass of another particle
$$d$$ is the distance So,

$$\dfrac{GM_0m}{x^2}=\dfrac{GM_02m}{(d-x)^2}$$

$$\dfrac{1}{x^2}=\dfrac{2}{(d-x)^2}$$

$$\dfrac{1}{x}=\dfrac{\sqrt{2}}{d-x}$$

$$x=\sqrt{2}=d-x$$

$$x(\sqrt{2}+1)=d$$

$$x=\dfrac{d}{(\sqrt{2}+1)}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}$$

$$x=\dfrac{(\sqrt{2}-1)d}{(\sqrt{2})^2-(1)^2}$$

$$\boxed{x=(\sqrt{2}-1)d}$$ from 2m
Question 8
1 / -0

Two coherent sources of wavelength $$6.2\times 10^{-7}$$m produce interference. The path difference corresponding to $$10^{th}$$ order maximum will be?

A
$$6.2\times 10^{-6}$$m

B
$$3.1\times 10^{-6}$$m

C
$$1.5\times 10^{-6}$$m

D
$$12.4\times 10^{-6}$$m

Solution

for $$ { m^{ th } }$$ order maximum path difference
$$ =n\lambda =10\times 6.2\times { 10^{ -7 } }\, m \\ =6.2\times { 10^{ -6 } } m$$
Option is A
Question 9
1 / -0

Unpolarised light of intensity $$I_{o}$$ passes through two polaroids; the axes of one is vertical. The intensity of transmitted light is:

A
$$\dfrac {l_{0}}{4}$$

B
$$\dfrac {l_{0}}{8}$$

C
$$\dfrac {l_{0}}{2}$$

D
$$\dfrac {3l_{0}}{4}$$
Question 10
1 / -0

In $$YDSE$$ interference pattern produced by two identical slits, the intensity at the site of the central maximum is $$I$$. the intensity at the same spot when either of two slits is closed is

A
$$0.5I$$

B
$$0.25I$$

C
$$2I$$

D
$$I$$

Solution

Let the amplitude of waves is A
Therefore,
$$ {A }_{ max }=A+A=2A$$
Now,
$${ I }_{ max }={ { A }_{ max } }^{ 2 }={ \left( 2A \right) }^{ 2 }=4A$$
Let intensity of each split is $${ I }_{ 0 }$$
Then,
When one split is closed
$${ I }_{ 0 }=\frac { { I }_{ max } }{ 4 } =0.25I$$
correct option is B.

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 49

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Wave Optics Test - 49

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SHARING IS CARING

Question 1

$$16.25$$

$$9:1$$

$$4:5$$

$$5:4$$

Solution

Question 2

$$1.53 \times 10^{11}$$

$$1.53 \times 10^{2}$$

$$1.53 \times 10^{4}$$

$$1.53 \times 10^{13}$$

Solution

Question 3

$$4$$

$$\dfrac{4}{3}$$

$$\dfrac{8}{3}$$

$$\dfrac{1}{4}$$

Solution

Question 4

$$25 : 1$$

$$5 : 1$$

$$9 : 4$$

$$4 : 9$$

Question 5

$${ cos }^{ -1 }\dfrac { 1 }{ \sqrt { 14 } } $$

$${ cos }^{ -1 }\dfrac { 2 }{ \sqrt { 14 } } $$

$${ sin }^{ -1 }\dfrac { 1 }{ \sqrt { 14 } } $$

$${ sin }^{ -1 }\dfrac { 2 }{ \sqrt { 14 } } $$

Solution

Question 6

$$6.8 \ mm$$

$$7.8 \ mm$$

$$9.8 \ mm$$

$$10.8 \ mm$$

Question 7

$$(\sqrt{2} + 1)d$$ from $$'2m'$$

$$(\sqrt{2} - 1)d$$ from $$'2m'$$

$$\dfrac{3d}{4}$$ from $$'m'$$

$$(\sqrt{2} - 1)d$$ from $$'m'$$

Solution

Question 8

$$6.2\times 10^{-6}$$m

$$3.1\times 10^{-6}$$m

$$1.5\times 10^{-6}$$m

$$12.4\times 10^{-6}$$m

Solution

Question 9

$$\dfrac {l_{0}}{4}$$

$$\dfrac {l_{0}}{8}$$

$$\dfrac {l_{0}}{2}$$

$$\dfrac {3l_{0}}{4}$$

Question 10

$$0.5I$$

$$0.25I$$

$$2I$$

$$I$$

Solution

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