Wave Optics Test

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Wave Optics Test - 70

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Question 1
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Directions For Questions

In YDSE setup, the light source executes SHM between $$P$$ and $$Q$$ according to the equation $$x=A\sin { \omega t } $$, $$S$$ being the mean position. Assume $$d\rightarrow 0$$ and $$A<<L$$. $$\omega$$ is small enough to neglect Doppler effect. If the source were stationary at $$S$$, intensity at $$O$$ would be $${I}_{0}$$.

...view full instructions

The fractional change in intensity of the central maximum as function of time is

A
$$\cfrac { A\sin { \omega t } }{ L } $$

B
$$\cfrac { 2A\sin { \omega t } }{ L } $$

C
$$\cfrac { 3A\sin { \omega t } }{ L } $$

D
$$\cfrac { 4A\sin { \omega t } }{ L } $$

Solution

$$I=2{ I }^{ 1 }(1+\cos { \delta } )$$
Where $$\delta =phase$$
$$\delta =\mu \dfrac { w }{ c } \triangle x$$
But, $$\triangle x=A \sin { wt } $$
$$\therefore \delta $$ is small thus,
$$I=\dfrac { 2A \sin { wt } }{ L } $$
Question 2
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Directions For Questions

In YDSE setup, the light source executes SHM between $$P$$ and $$Q$$ according to the equation $$x=A\sin { \omega t } $$, $$S$$ being the mean position. Assume $$d\rightarrow 0$$ and $$A<<L$$. $$\omega$$ is small enough to neglect Doppler effect. If the source were stationary at $$S$$, intensity at $$O$$ would be $${I}_{0}$$.

...view full instructions

When the source comes toward the point $$Q$$,

A
the bright fringes will be less bright

B
the dark fringes will no longer remain dark

C
the fringe width will increase

D
none of these

Solution

As the Source is approaching towards slit setup,
the number of Photons hitting the screen will be more than the situation when the Source is stationary.

Hence the Fringes will become Brighter with no change in Fringe Width

Answer is D: None of These
Question 3
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Directions For Questions

The arrangement for a mirror experiment is shown in figure. $$'S'$$ is a point source of frequency $$6\times {10}^{14}Hz$$. $$D$$ and $$C$$ represent the two ends of a mirror placed horizontally and $$LOM$$ represents the screen.

...view full instructions

Find the fringe width of the fringe pattern.

A
$$0.05cm$$

B
$$0.25cm$$

C
$$0.01cm$$

D
$$0.1cm$$

Solution

$$S'$$ is the virtual point source such that light from $$S$$ and $$S'$$ interfere on the the screen, then this situation resembles with the Young's double slit experiment.
Let d be the distance between the two point source.
$$d=2 SP= 2 mm$$
$$D=2m v=3 \times 10^8 ms^{-1} \nu = 6\times 10^{14} Hz$$
using $$v = \nu \lambda$$
On solving, $$\lambda = 500 nm$$
fringe width, $$\beta= \dfrac{\lambda D}{d}$$
This gives $$\beta = .05 cm$$
Question 4
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Directions For Questions

The arrangement for a mirror experiment is shown in figure. $$'S'$$ is a point source of frequency $$6\times {10}^{14}Hz$$. $$D$$ and $$C$$ represent the two ends of a mirror placed horizontally and $$LOM$$ represents the screen.

...view full instructions

Calculate the number of fringes.

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

As calculated above, $$\beta= .05 cm $$
Region for which fringes are visible, $$y = 2 cm$$
Number of fringes, $$N = \dfrac{y}{\beta}$$
This gives $$N = 40$$
Question 5
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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. If the intensity at $$p$$ be maximum, then the minimum distance through which the reflecting surface $$MN$$ should be displaced so that at $$P$$ again becomes maximum?

A
3 $$\times$$ 10$$^{-7}$$ m

B
6 $$\times$$ 10$$^{-7}$$ m

C
1.5 $$\times$$ 10$$^{-7}$$ m

D
12 $$\times$$ 10$$^{-7}$$ m

Solution

Light reaches to point P by two ways.
First, directly from source S and second, after reflection from mirror MN. Therefore path difference,
$$\Delta x=2h$$,
Now, let the surface MN is displaced by $$y$$.
Hence the path difference,
$$\Delta x'=2(h\pm y)$$ ,
Extra path difference due to displacement of surface MN,
$$\Delta x''=2(h\pm y)-2h=\pm2 y$$
The intensity of light at P will be maximum again, if
$$2y=\lambda=6000$$,
or $$y=6000/2=3000A^{o}$$,
or $$y=3\times10^{-7}m$$
Question 6
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A distant star which is moving away with a velocity of $$ 10^6$$ m/sec is emitting red line of frequency$$4.5\times 10^14 $$ Hz. The observed frequency of this spectral line is

A
$$ 4.5\times 10^8 Hz $$

B
$$ 4.485\times 10^{14} Hz $$

C
$$ 4.515\times 10^{14} Hz $$

D
$$ 4.5\times 10^{14} Hz $$

Solution

The source is moving with a velocity, $$u = 10^6\ m/s$$
The frequency will be Doppler shifted.
The relationship between observed frequency ($$\nu'$$) and emitted frequency ($$\nu_0$$) is given by: $$\nu' = \dfrac{v+u_r}{v+u_s}\nu_0$$
where, $$v$$ is the velocity of waves in the medium.
$$u_r$$ is the velocity of the receiver relative to the medium; positive if the receiver is moving towards the source (and negative in the other direction);
$$u_s$$ is the velocity of the source relative to the medium; positive if the source is moving away from the receiver (and negative in the other direction).

Here $$u_r = 0$$ and $$u_s = 10^6$$ $$m/s$$
$$\nu' = \dfrac{3 \times 10^8}{3 \times 10^8+10^6}4.5 \times 10^{14}=4.485 \times 10^{14}$$ $$Hz$$
Question 7
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Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is :

A
$$1\, \mu m$$

B
$$30\, \mu m$$

C
$$100\, \mu m$$

D
$$300\, \mu m$$

Solution

Diameter $$d = 2r = 2 \times 0.25 = 0.5 cm$$, the wavelength of light is $$\lambda= 500 nm = 5 \times 10^{-7} m$$.

We have the formula $$sin\theta=\dfrac{1.22\lambda}{d}$$
or
$$sin\theta=\dfrac{1.22\times 5\times 10^{-7}}{0.5\times 10^{-2}}=1.22\times 10^{-4}$$

Also $$D=25cm$$

Let $$x$$ be the minimum separation between two objects that the human
eye can resolve.

Thus $$sin\theta=\dfrac{x}{D}$$
or
$$x=Dsin\theta=.25\times 1.22\times10^{-4}=30\times 10^{-6}=30\mu m$$
Question 8
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Two identical coherent sources are placed on a diameter of a circle of radius R at a separation d (d << R, d >> $$\lambda$$) symmetrically about the centre of the circle. The sources emit identical wavelength $$\lambda$$. The number of the points on the circle with maximum intensity is

A
$$\displaystyle \frac{2d}{\lambda }+1$$

B
$$\displaystyle \frac{4d}{\lambda }$$

C
$$\displaystyle \frac{4d}{\lambda }-2$$

D
$$\displaystyle \frac{4d}{\lambda }+2$$

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 $$ n=\dfrac { 4d }{ \lambda } $$
Question 9
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Light is incident at an angle $$\displaystyle \phi $$ with the normal to a plane containing two slits of separation d Select the expression that correctly describe the positions of the interference maxima in terms of the incoming angle $$\displaystyle \theta $$ and outgoing $$\displaystyle \phi $$

A
$$\displaystyle \sin \phi \sin \theta =\left ( m+\frac{1}{2} \right )\frac{\lambda }{d}$$

B
$$\displaystyle d\sin \theta =m\lambda $$

C
$$\displaystyle \sin \phi -\sin \theta (m+1)\frac{\lambda }{d}$$

D
$$\displaystyle \sin \phi +\sin \theta =m\frac{\lambda }{d}$$

Solution
Question 10
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A paper, with two marks having separation d, is held normal to the line of sight of an observer at a distance of 50m. The diameter of the eye-lens of the observer is 2 mm. Which of the following is the least value of d, so that the marks can be seen as separate? (The mean wavelength of visible light may be taken as 5000 $$A^o$$)

A
$$1.25 m$$

B
$$12.5 cm$$

C
$$1.25 cm$$

D
$$2.5 mm$$

Solution

Angular limit of resolution of eye, $$\theta = \displaystyle \frac{\lambda}{d}$$, where d is diameter of eye lens.
Also if y is the minimum just resolution separation between two objects at distance D from eye then $$\displaystyle \theta = \frac{y}{D}$$
$$\displaystyle \Rightarrow \frac{y}{D}= \frac{\lambda}{d} \Rightarrow y = \frac{\lambda D}{d}$$ ............ (1)
Here: wavelength $$=\lambda = 5000 A^o = 5 \times 10^{-7} m D = 50 m$$
Diameter of eye lens $$= 2mm = 2 \times 10^{-3} m$$
From equation (1) minimum separation is
$$\displaystyle y = \frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}} = 12.5 \times 10^{-3} m = 12.5 cm$$