Wave Optics Test

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Wave Optics Test - 42

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Question 1
1 / -0

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern.

A
The intensities of both the maxima and the minima increase

B
The intensity of the maxima increases and the minima has zero intensity

C
The intensity of the maxima decrease and that of the minima increases

D
The intensity of the maxima decreases and the minima has zero intensity

Solution

The intensities due to slits are proportional to their width.
When we takes slits of equal width
$$\dfrac {I_{1}}{I_{2}} = \dfrac {W_{1}}{W_{2}} = 1$$
$$\dfrac {I_{1}}{I_{2}} = \dfrac {a^{2}}{b^{2}} = 1$$
or $$\dfrac {a}{b} = 1$$
$$\dfrac {I_{max}}{I_{min}} = \dfrac {(a + b)^{2}}{(a - b)^{2}} = \dfrac {\left (\dfrac {a}{b} + 1\right )^{2}}{\left (\dfrac {a}{b} - 1\right )^{2}} = \dfrac {(1 + 1)^{2}}{(1 - 1)^{2}} = \dfrac {2}{0}$$
This implies maximum intensity $$I_{max}$$ is $$2$$ and minimum intensity $$I_{min}$$ is zero.
When we take the width of one slit is twice of the other
$$\dfrac {I_{1}'}{I_{2}'} = \dfrac {W_{1}'}{W_{2}'} = 2$$
$$\dfrac {I_{1}'}{I_{2}'} = \dfrac {a'^{2}}{b'^{2}} = \dfrac {2}{1}$$
or $$\dfrac {a'}{b'} = \dfrac {\sqrt {2}}{1}$$
$$\dfrac {I_{max}'}{I_{min}'} = \dfrac {(a' + b')^{2}}{(a' - b')^{2}} = \dfrac {\left (\dfrac {a'}{b'} + 1\right )^{2}}{\left (\dfrac {a'}{b'} - 1\right )^{2}} = \dfrac {(\sqrt {2} + 1)^{2}}{(\sqrt {2} - 1)^{2}}$$.
This implies maximum intensity $$I'_{max}$$ is $$(\sqrt{2}+1)^2$$ and minimum intensity $$I_{min}$$ is $$(\sqrt{2} -1)^2$$.
Therefore, the intensities of both the maxima and minima increase.
Question 2
1 / -0

A fixed source emits sound of frequency 1000 Hz. What is the frequency as heard by a observer at rest?

A
$$900 Hz$$

B
$$1000 Hz$$

C
$$1080 Hz$$

D
$$700 Hz$$

Solution

Doppler effect is the apparent change in frequency of sound when the source, observer and medium are in the relative motion. As here the both source and observer are in rest, so we can not observe Doppler effect of sound. Thus, there is no change in frequency of sound as heard by observer. Therefore, the frequency of the sound as heard by observer will be the frequency of the source i.e. $$1000 $$ Hz.
Question 3
1 / -0

Which of the following principles is being used in Sonar Technology?

A
Newton's laws of motion

B
Reflection of electromagnetic waves

C
Laws of thermodynamics

D
Reflection of ultrasonic waves

Solution

sonar technology uses Doppler effect of ultrasonic waves
Question 4
1 / -0

In Doppler effect when the source is moving away from the observer at rest, the correct equation is:

A
$$f'=f(\dfrac{v-v_s}{v+v_s})$$

B
$$f'=f(\dfrac{v_s}{v+v_s})$$

C
$$f'=f(\dfrac{v}{v+v_s})$$

D
None of the above

Solution

Source moving away from observer:
$$f'=f\left(\dfrac {v_{wave}}{v_{wave}+v_s}\right)$$
Question 5
1 / -0

If Young's double slit experiment is done with white light, which of the following statements will be true?

A
All the bright fringes will be coloured

B
All the bright fringes will be white

C
The central fringe will be white

D
No stable interference pattern will be white

Solution

$$\Delta x=0$$ at centre for all wavelengths.
Question 6
1 / -0

In a Young's double slit experiment the intensity of the resultant wave at a point P on the screen is I where the path difference between the waves from coherent sources $$S_1$$ and $$S_2$$ is $$\lambda$$. Then the intensity of the resultant wave at a point where the path difference is $$\lambda /4$$ is given by.

A
$$I/\sqrt{2}$$

B
$$2I$$

C
$$4I$$

D
$$I/2$$

Solution

Given,
The intensity of the resultant wave at a point P on the screen is $$I$$ when the path difference is $$\lambda$$
Here we have to find the intensity of the resultant wave at a point where the path difference is $$\lambda/4$$.
We have Intensity, $$I = I $$ $$\cos^2\dfrac\phi2$$ ............... (1)
Where $$\phi$$ is phase difference
Also, Phase difference, $$\phi$$ $$ = \dfrac {2\pi}{\lambda}$$ X path difference .................(2)
Given, path difference is $$\dfrac{\lambda}{4}$$
So, pahse difference, $$ \phi$$ = $$\dfrac{2\pi}{\lambda}$$ X $$\dfrac\lambda4$$
That is, $$\phi$$ = $$\dfrac\pi2$$
Substituting the value of $$\phi$$ in equation (1) we get,
Intensity, $$I$$ = $$I$$ $$\cos^2\dfrac\pi4$$
Question 7
1 / -0

If the wavelength of light used is $$6000\mathring { A } $$. The angular resolution of telescope of objective lens having diameter $$10cm$$ is ______ rad

A
$$7.52\times { 10 }^{ -6 }$$

B
$$6.10\times { 10 }^{ -6 }$$

C
$$6.55\times { 10 }^{ -6 }$$

D
$$7.32\times { 10 }^{ -6 }$$

Solution

Limit of resolution $$\sin { \theta } =\theta =\cfrac { 1.22\lambda }{ D } $$
putting the values
$$\theta=\dfrac{1.22\times6000\times10^{-10}}{0.1}$$

$$\theta=7.32\times10^{-6}$$

Option (D) is correct.
Question 8
1 / -0

With the help of a telescope that has an objective of diameter $$200cm$$, it is proved that light of wavelengths of the order of $$6400\mathring { A } $$ coming from a star can be easily resolved. Then the limit of resolution is

A
$$39\times { 10 }^{ -8 }deg$$

B
$$39\times { 10 }^{ -8 }rad$$

C
$$19.5\times { 10 }^{ -8 }rad$$

D
$$19.5\times { 10 }^{ -8 }deg$$

Solution

Given : $$D = 200 \ cm = 2 \ m$$ $$\lambda = 6400\times 10^{-10} \ m = 6.4\times 10^{-7} \ m$$
Limit of resolution of telescope $$d\theta = \dfrac{1.22\lambda}{D}$$
$$\implies \ d\theta = \dfrac{1.22\times 6.4\times 10^{-7}}{2} = 39\times 10^{-8} \ rad$$
Question 9
1 / -0

Two coherent point sources of sound wave $$S_1$$ and $$S_2$$ produce sound of same frequency 50 Hz and wavelength 2 cm with amplitude 2 x $$(10)^-$$$$^3$$ m. Each circular arc represents a wavefront at a particular time and is separated from next arc by a distance 1 cm. Both the sound waves propagate through the medium and interfere with each other. Read paragraph carefully and answer the following questions. [r = 1 cm]
The point (s) where constructive interference occurs

A
G only

B
P and A

C
G and F

D
T and U
Question 10
1 / -0

Monochromatic light of wavelengths $$400$$ nm and $$560$$ nm are incident simultaneously and normally on double-slit apparatus whose slit separation is $$0.1$$ mm and screen distance is $$1 m$$. Distance between areas of total darkness will be ___ mm

A
$$4 $$

B
$$5.6 $$

C
$$14 $$

D
$$28$$

Solution

Given:
Wavelengths, $$\lambda_1=400nm=400\times10^{-9}m$$,
$$\lambda_2=560nm=560\times10^{-9}m$$
Slit separation, $$d=0.1mm=0.1\times 10^{-3}m$$
Screen distance, $$D=1m$$
To find:
Distance between areas of total darkness, $$\Delta s=?$$
According to the question,
$$(2n+1)\lambda_1=(2m+1)\lambda_2\\\implies \dfrac {(2n+1)}{(2m+1)}=\dfrac {\lambda_2}{\lambda_1}=\dfrac {560nm}{400nm}=\dfrac {7}{5}$$

$$10n+5=14m+7\\\implies 10n=14m+2\\\implies 5n=7m+1$$

By inspection, for $$m_1=2, n_1=3$$ and $$m_2=7, n_2=10$$

Therefore, $$\Delta s=\dfrac {\lambda_1 D}{2d}[(2n_2+1)-(2n+1)]$$
$$=\dfrac {400\times 10^{-9}\times1}{2\times0.1\times 10^{-3}}[(2\times10+1)-(2\times3+1)]$$

$$=200\times10^{-5}(21-7)$$

$$\implies \Delta s=2800\times10^{-5}=28mm $$
Hence, the distance between areas of total darkness will be $$28 mm$$

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Wave Optics Test - 42

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Wave Optics Test - 42

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Question 1

The intensities of both the maxima and the minima increase

The intensity of the maxima increases and the minima has zero intensity

The intensity of the maxima decrease and that of the minima increases

The intensity of the maxima decreases and the minima has zero intensity

Solution

Question 2

$$900 Hz$$

$$1000 Hz$$

$$1080 Hz$$

$$700 Hz$$

Solution

Question 3

Newton's laws of motion

Reflection of electromagnetic waves

Laws of thermodynamics

Reflection of ultrasonic waves

Solution

Question 4

$$f'=f(\dfrac{v-v_s}{v+v_s})$$

$$f'=f(\dfrac{v_s}{v+v_s})$$

$$f'=f(\dfrac{v}{v+v_s})$$

None of the above

Solution

Question 5

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 white

Solution

Question 6

$$I/\sqrt{2}$$

$$2I$$

$$4I$$

$$I/2$$

Solution

Question 7

$$7.52\times { 10 }^{ -6 }$$

$$6.10\times { 10 }^{ -6 }$$

$$6.55\times { 10 }^{ -6 }$$

$$7.32\times { 10 }^{ -6 }$$

Solution

Question 8

$$39\times { 10 }^{ -8 }deg$$

$$39\times { 10 }^{ -8 }rad$$

$$19.5\times { 10 }^{ -8 }rad$$

$$19.5\times { 10 }^{ -8 }deg$$

Solution

Question 9

G only

P and A

G and F

T and U

Question 10

$$4 $$

$$5.6 $$

$$14 $$

$$28$$

Solution

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