Wave Optics Test

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Wave Optics Test - 71

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Question 1
1 / -0

Directions For Questions

A point source is emitting light of wavelength 6000 $$\overset{o}{A}$$ is placed at a very small height h above a flat reflecting surface $$MN$$ as shown in the figure. The intensity of the reflected light is $$36$$% of the incident intensity. Inference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $$D$$ from it.

...view full instructions

Ratio of maximum to minimum intensities at $$P$$ is

A
$$2 : 1$$

B
$$4 :1$$

C
$$8 : 1$$

D
$$16 : 1$$

Solution

Before reflection, intensity of light is $$ {I}_{o}$$(say) and after reflection it becomes $$ {0.36I}_{o}$$
$$ { I }_{ max }={ I }_{ o }+{ 0.36I }_{ o }+2\sqrt { { I }_{ o }{ 0.36I }_{ o } } =2.56{ I }_{ o }$$
$$ { I }_{ min }={ I }_{ o }+{ 0.36I }_{ o }-2\sqrt { { I }_{ o }{ 0.36I }_{ o } } =0.16{I}_{o}$$
$$\dfrac { { I }_{ max } }{ { I }_{ min } } =\dfrac { 256 }{ 16 } =16:1$$
Question 2
1 / -0

Directions For Questions

A point source is emitting light of wavelength 6000 $$\overset{o}{A}$$ is placed at a very small height h above a flat reflecting surface $$MN$$ as shown in the figure. The intensity of the reflected light is $$36$$% of the incident intensity. Inference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $$D$$ from it.

...view full instructions

The shape of the interference fringes, on the screen is

A
circle

B
ellipse

C
parabola

D
straight line

Solution

$$S'$$ represents the another point source forms due to reflection from the mirror.
$$S'P= \sqrt{(x+h)^2+ y^2} $$ and $$SP= \sqrt{(x-h)^2+ y^2} $$
Let path difference at point P be $$\Delta$$
Using, $$\Delta + SP= S'P$$
Squaring and adding, $$\Delta^2 + 4hx= -2\Delta\sqrt{(x-h)^2+ y^2}$$
Squaring and adding again, $$\Delta^4 + 4h^2x^2 + 8hx\Delta= 4\Delta^2(x^2 +h^2- 2hx + y^2)$$
$$\implies 4(\Delta^2-h^2)x^2 - (8\Delta^2h+8h\Delta)x + y^2+ 4\Delta^2h^2-\Delta^4= 0$$
which represents equation of circle as $$X^2 + Y^2= R^2$$
Thus the fringes will of Circular shape centered at O.
Question 3
1 / -0

In Young's double slit experiment, if the widths of the slit are in the ratio $$4:9$$ , ratio of intensity of maxima to intensity of minima will be

A
$$25:1$$

B
$$9:4$$

C
$$3:2$$

D
$$81:16$$

Solution

$$\textbf{HINT:}$$ the measurable amount of a property, such as force, brightness, or a magnetic field.

$$\textbf{Step1:}$$width of silt is directly proptiontal to intensity of light

Width of slit $$(\mathrm{W}) \propto$$ Light intensity (I),$$\dfrac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\dfrac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\dfrac{4}{9}=$$ $$\dfrac{4 \mathrm{k}}{9 \mathrm{k}} \quad \ldots(\mathrm{k}$$ is a constant $$)$$

$$I_{\max }=4 \mathrm{k}+9 \mathrm{k}+2 \sqrt{4 \mathrm{k} 9 \mathrm{k}}=25 \mathrm{k} \\$$

$$\mathrm{I}_{\min }=4 \mathrm{k}+9 \mathrm{k}-2 \sqrt{4 \mathrm{k} 9 \mathrm{k}}=\mathrm{k} \\$$

$$\textbf{Step 2:}$$ Dividing the expression,
$$\dfrac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=25:1$$
Question 4
1 / -0

In Young's double slit experiment, if monochromatic light used is replaced by white light then :

A
no fringes are observed

B
only central fringe is white, all other fringes are coloured.

C
all bright fringes become white.

D
all bright fringes have colour between violet and red.

Solution

The difference in path lengths, being zero at the center gives zero phase difference at the center for all wavelengths. Hence the center fringe will be white. But at other points on the screen, phase difference will be given by $$\phi =k\Delta x=\dfrac { 2\pi }{ \lambda } \Delta x$$ where $$\Delta x$$ is the difference in path lengths. Since $$\lambda$$ is different for different colours, hence fringe width will be different for each colour and hence all other fringes will be coloured.
Question 5
1 / -0

When an object is viewed with a light of wavelength $$6000\mathring{A}$$ under a microscope, its resolving power is $$10^4$$. The resolving power of the microscope when the same object is viewed with a light of wavelength $$4000\mathring{A}$$, is:

A
$$1.5 \times 10^4$$

B
$$2\times 10^4$$

C
$$3\sqrt{2}\times 10^4$$

D
$$3\times 10^4$$

E
$$ 10^4$$

Solution

The resolving power of microscope is given by the following formula.
$$R=\displaystyle\frac{2\mu \sin\theta}{1.22\lambda}$$
where, $$\mu =$$ refractive index
$$\lambda =$$ wavelength of the height
$$\therefore R\propto \displaystyle\frac{1}{\lambda}$$
$$\displaystyle\frac{R_2}{R_1}=\frac{\lambda_1}{\lambda_2}=\frac{6000}{4000}\overset{o}{A}$$
$$\displaystyle R_2=\frac{3}{2}\times R_1=\frac{3}{2}\times 10^4$$
$$=1.5\times 10^4$$.
Question 6
1 / -0

A: The corpuscular theory fails in explaining the velocities of light in air and water.
B: According to corpuscular theory, the light should travel faster in a 'denser medium than in a rarer medium.

A
If both A and B are true but the B is the correct explanation of A

B
If both A and B are true but the B is not the correct explanation of A

C
If A is true but B is false

D
If both the A and B are false

E
If B is true but A is false

Solution

Corpuscular theory fails to explain the velocity of light in air and water because it predicted light to have more velocity in denser medium where as the fact is just the opposite. So option (a) is correct.
Question 7
1 / -0

Directions For Questions

The figure shows a surface XY separating two transparent media, medium - 1 and medium - 2. The lines ab and cd represent wavefronts of a light wave travelling in medium - 1 and incident on XY. The lines ef and gh represent wavefronts of light wave in medium - 2 after refraction.

...view full instructions

The phases of the light wave at c, d, e and f are $$\phi_c, \phi_d, \phi_e$$ and $$\phi_f$$ respectively. It is given that $$\phi_c \neq \phi_r$$

A
$$\phi_c$$ cannot be equal to $$\phi_d$$

B
$$\phi_d$$ cannot equal to $$\phi_e$$

C
$$(\phi_d - \phi_f)$$ is equal to $$(\phi_c - \phi_e)$$

D
$$(\phi_d - \phi_c)$$ is not equal to $$(\phi_f - \phi_e)$$

Solution

All the point on wavefront are in same phase of oscillation. Therefore
$$\phi_c=\phi_d$$ and $$\phi_f=\phi_e$$
So $$\phi_d-\phi_f=\phi_c-\phi_e$$
Question 8
1 / -0

Light of wavelength $$6000 \mathring {A}$$ is reflected at nearly normal incidence from a soap film of refractive index 1.4. The least thickness of the film that will appear black is.

A
Infinity

B
$$200 \mathring{A}$$

C
$$2000 \mathring{A}$$

D
$$1000 \mathring{A}$$

Solution

Interference occurs between two reflected rays A and B, A is reflected from upper surface and B is reflected from lower surface.
The path difference between the two becomes $$2\mu t +\dfrac{\lambda}{2}$$
If it is equal to $$\displaystyle \dfrac{3\lambda}{2}$$ then destructive interference occurs and we see dark fringes from above. So, the equation $$\displaystyle 2\mu t +\dfrac{\lambda}{2} = \dfrac{3\lambda}{2}$$ gives the least value oft for which we see dark fringes.
$$\displaystyle 2\mu t = \lambda \Rightarrow t = \dfrac{\lambda}{2\mu} = \dfrac{6000}{2\times 1.4} \cong 2000\mathring{A}$$
Question 9
1 / -0

A plane wave of monochromatic light falls normally on a uniformly thin film of oil which covers a glass plate. The wavelength of source constructive interference is observed for $$\lambda_{1} = 5000\overset {\circ}{A}$$ and $$\lambda_{2} = 10000\overset {\circ}{A}$$ and for no other wavelength in between. If $$\mu$$ of oil is $$1.25$$ and that of glass is $$1.5$$, the thickness of film will be ________ $$\mu m$$.

A
$$0.2$$

B
$$0.1$$

C
$$0.8$$

D
$$0.4$$

Solution

Given:
$$\mu_o=1.25, \mu_{g}=1.5, \lambda_1 = 5 \times 10^{-7}m=\lambda_{n+1}$$ (No wavelength between destructive interference)
$$\lambda_2=10\times10^{-7}m$$
The set u is similar to anti-reflective coating.
For destructive interference
$$2\mu \cos r=(2n-1)\dfrac {\lambda_n}2=[2(n+1)-1]\dfrac {\lambda_{n+1}}2......(i)$$
$$(2n-1)\dfrac {5\times 10^{-7}}{2}=(2n+1)\dfrac {10\times10^{-7}}{2}$$
$$n=\dfrac 32$$....(ii)
For, normal incidence, $$cos r =1,$$
from (i) and (ii)
$$2\mu t=\left(2\times \dfrac 32-1\right)\dfrac {\lambda_n}{2}=\lambda_n=5\times 10^{-7}$$
$$t=\dfrac {5\times 10^{-7}}{2\times1.25}=2\times 10^{-7}m$$
or the thickness = $$0.2\mu m$$
Question 10
1 / -0

Two cohorent monochromatic light beams of intensities 4I and 9I interfere in a Young's double slit experimental setup to produce a fringe pattern on the screen. The phase difference between the beams at two points P and Q on the screen are $$\pi/2$$ and $$\pi/3$$ respectively. Then the ratio of the two intensities $$I_P/I_Q$$ is

A
0

B
$$\displaystyle \frac{6}{19}$$

C
$$\displaystyle \frac{13}{19}$$

D
$$\displaystyle \frac{6}{13}$$

Solution

We know that $$I \propto A^2 $$ where $$I$$ is the intensity and $$A$$ is the amplitude.
Given that $$I_1=4I$$ and $$I_2=9I$$
Hence,
$$\dfrac{A_1}{A_2}=\sqrt{\dfrac{I_1}{I_2}}$$

$$\implies\dfrac{A_1}{A_2}=\sqrt{\dfrac{4I}{9I}}$$

$$\implies\dfrac{A_1}{A_2}=\dfrac{2}{3}$$

Let, $$A_1=2A$$ and $$A_2=3A$$

When, phase difference$$(\phi)$$ is $$\dfrac{\pi}{2}$$ :-

Amplitude $$A_P=\sqrt{A_1^2+A_2^2+2A_1A_2\cos{\phi}}$$

$$\implies A_P=\sqrt{4A^2+9A^2+0}=\sqrt{13}A$$

When, phase difference$$(\phi)$$ is $$\dfrac{\pi}{3}$$ :-

Amplitude $$A_Q=\sqrt{A_1^2+A_2^2+2A_1A_2\cos{\phi}}$$

$$\implies A_Q=\sqrt{4A^2+9A^2+2.2A.3A.\dfrac{1}{2}}=\sqrt{19}A$$

Now, $$\dfrac{I_P}{I_Q}=\dfrac{A_P^2}{A_Q^2}$$

$$=\dfrac{13}{19}$$

Hence, answer is option-(C).

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 71

Result

Wave Optics Test - 71

Score

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SHARING IS CARING

Question 1

$$2 : 1$$

$$4 :1$$

$$8 : 1$$

$$16 : 1$$

Solution

Question 2

circle

ellipse

parabola

straight line

Solution

Question 3

$$25:1$$

$$9:4$$

$$3:2$$

$$81:16$$

Solution

Question 4

no fringes are observed

only central fringe is white, all other fringes are coloured.

all bright fringes become white.

all bright fringes have colour between violet and red.

Solution

Question 5

$$1.5 \times 10^4$$

$$2\times 10^4$$

$$3\sqrt{2}\times 10^4$$

$$3\times 10^4$$

$$ 10^4$$

Solution

Question 6

If both A and B are true but the B is the correct explanation of A

If both A and B are true but the B is not the correct explanation of A

If A is true but B is false

If both the A and B are false

If B is true but A is false

Solution

Question 7

$$\phi_c$$ cannot be equal to $$\phi_d$$

$$\phi_d$$ cannot equal to $$\phi_e$$

$$(\phi_d - \phi_f)$$ is equal to $$(\phi_c - \phi_e)$$

$$(\phi_d - \phi_c)$$ is not equal to $$(\phi_f - \phi_e)$$

Solution

Question 8

Infinity

$$200 \mathring{A}$$

$$2000 \mathring{A}$$

$$1000 \mathring{A}$$

Solution

Question 9

$$0.2$$

$$0.1$$

$$0.8$$

$$0.4$$

Solution

Question 10

0

$$\displaystyle \frac{6}{19}$$

$$\displaystyle \frac{13}{19}$$

$$\displaystyle \frac{6}{13}$$

Solution

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