Wave Optics Test

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Wave Optics Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

A light source approaches the observer with velocity 0.8C. The Doppler shift of or light of the wavelength $$5500\mathop {\rm{A}}\limits^{\rm{0}} $$ is:

A
$$4400\mathop {\rm{A}}\limits^{\rm{0}} $$

B
$$1833\mathop {\rm{A}}\limits^{\rm{0}} $$

C
$$3167\mathop {\rm{A}}\limits^{\rm{0}} $$

D
$$7333\mathop {\rm{A}}\limits^{\rm{0}} $$

Solution

According to Doppler's Principle
$$\lambda=\lambda\sqrt{\cfrac{1-V/C}{1+V/C}}$$ for $$V=C$$
$$\lambda=5500\sqrt{\cfrac{1-0.8}{1+0.8}}=1833.3\\ \therefore Shift=55001833.3=3167AA$$
Question 2
1 / -0

In a young's double slit experiment , the distance between the two slits is 0.1mm and the wavelength of light used is 4$$\times 10^{-7}$$ m . If the width of the fringe on the screen is 4mm, then the distance between screen and slit is

A
0.1mm

B
1cm

C
0.1 cm

D
1m

Solution
Question 3
1 / -0

A photograph of the moon was taken with telescope. Later on, it was found that a housefly was siting on the objective lens of the telescope. In photograph

A
the image of the housefly will be reduced

B
there is a reduction in the intensity of the image

C
there is an increase in the intensity of the image

D
the image of the housefly will be enlarged

Solution

A photograph of the moon was taken with a telescope. Later on, it was found that a housefly was sitting on the objective lens of the telescope. In the photograph the image of the housefly will be enlarged because size of the aperture decreases.
Question 4
1 / -0

The interference pattern with two coherent light sources of density ratio $$n$$. In the interference pattern, the ratio $$\dfrac {I_{max}-I_{min}}{ I_{max}+I_{min}}$$ will be:

A
$$\dfrac {\sqrt {n}}{n+1}$$

B
$$\dfrac {2 \sqrt {n}}{n+1}$$

C
$$\dfrac {\sqrt {n}}{(n+1)^{2}}$$

D
$$\dfrac {2 \sqrt {n}}{(n+1)^{2}}$$

Solution

Given that $$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =n\\ { I }_{ max }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\ { I }_{ min }={ I }_{ 1 }+{ I }_{ 2 }-2\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\ \therefore \cfrac { { I }_{ max }+{ I }_{ min } }{ { I }_{ max }-{ I }_{ min } } =\cfrac { 4\sqrt { { I }_{ 1 }{ I }_{ 2 } } }{ 2\sqrt { { I }_{ 1 }{ I }_{ 2 } } } \\ \quad \quad \quad \quad \quad \quad =\cfrac { 4{ I }_{ 2 }\sqrt { { I }_{ 1 }/{ I }_{ 2 } } }{ 2{ I }_{ 2 }\sqrt { { I }_{ 1 }/{ I }_{ 2 }+1 } } \\ \quad \quad \quad \quad \quad \quad =\cfrac { 2\sqrt { n } }{ n+1 } \\ $$
Question 5
1 / -0

In a double slit experiment, the separation between the slits is d = 0.25 cm and the distance of the screen D = 100 cm from the slits, If the wavelength of light used is $$\lambda = 6000\mathop {\text{A}}\limits^{\text{o}} \,{\text{and}}\,{{\text{I}}_{\text{0}}}$$ is the intensity of the central bright fringe, the intensity at a distance $$y = 4 \times {10^{ - 5}}$$ m from the central maximum is

A
$${I_0}$$

B
$${I_0}$$/2

C
3$${I_0}$$/4

D
$${I_0}$$/3

Solution

It is given that,

$$ d=0.25\,cm $$

$$ D=100\,cm $$

$$ \lambda =6000\,\dot{A}=6\times {{10}^{-7}}\,m $$

For central maximum

$$ x=\dfrac{Dn\lambda }{d} $$

$$ x=\dfrac{100\times 6\times {{10}^{-7}}}{0.25} $$

$$ x=4\times {{10}^{-5}} $$

Intensity at this point is $${{I}_{0}}$$

At $$y=4\times {{10}^{-5}}\,m$$

Intensity will be $${{I}_{0}}$$
Question 6
1 / -0

Ray optics is valid when characteristic dimension are

A
much smaller than wavelength of light.

B
of same order as wavelength of light

C
much larger than wavelength of light

D
none of the above

Solution

Ray optics is valid when characteristic dimension are much larger than wavelength of light . The size of obstacle must be much larger than the wavelength of light. If wavelength is comparable to the size of object, then diffraction could happen, but it cannot be explained using ray optics, it will require wave theory of light.
Question 7
1 / -0

A broad source of light (l = 680 nm) illuminates normally two glass plates 120 mm long that touch at end and are separated by a wire 0.048 mm in diameter at the other end. The total number of bright fringes that appear over the 120 mm distance is:

A
50

B
100

C
141

D
400

Solution

Fringe Width $$\beta$$ in case of variable thickness wedge shaped film is ;
$$\beta = \frac{\lambda }{{2\mu \theta }}$$
we get $${\mathop{\rm Tan}\nolimits} \theta = \theta = \frac{y}{I}............\left( 1 \right)$$
Let $$N$$ be the number of gringes formed over length $$I$$
$$\begin{array}{l} I=N\beta \\ I=N\left( { \frac { \lambda }{ { 2\mu \theta } } } \right) \\ N=\frac { { 2\mu \theta } }{ \lambda } \\ N=\frac { { 2uY } }{ \lambda } \, \, \left[ { \because \theta =\frac { y }{ I } } \right] \\ N=\frac { { 2\times 1\times 0.048\times { { 10 }^{ -2 } } } }{ { 680\times { { 10 }^{ 9 } } } } \\ N=\frac { { 0.096\times { { 10 }^{ -2 } } } }{ { 680\times { { 10 }^{ 9 } } } } \\ N=141 \end{array}$$
Question 8
1 / -0

Light of wavelength $$\lambda $$ from a point source falls on a small circular obstacle of diameter d. Dark and bright circular rings around a central bright spot are formed on a screen beyond the obstacle. The distance between the screen and obstacle is D. Then , the condition for the formation of rings, is

A
$$\sqrt { \lambda } \approx \cfrac { { d } }{ 4D } $$

B
$$\lambda \approx \cfrac { { d }^{ 2 } }{ 4D } $$

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

D
$$\lambda \approx \cfrac { D }{ 4 } $$

Solution
Question 9
1 / -0

In Young's double slit experiment, the ratio of intensities of bright and dark frings is $$9$$. This means that

A
the intensities of individual source are $$5$$ and $$4$$ units respectively

B
the intensities of individual source are $$4$$ and $$1$$ units respectively

C
the ratio of their amplitude is $$3$$

D
the ratio of their amplitude is $$4$$

Solution

$$ \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}=\dfrac{9}{1} $$

$$ \dfrac{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}=\dfrac{3}{1} $$

$$ \sqrt{\dfrac{{{I}_{1}}}{{{I}_{2}}}}=\dfrac{2}{1} $$

$$ \dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{4}{1} $$

So, the ratios of their intensities is $$4:1$$
Question 10
1 / -0

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

A
$$169:25$$

B
$$81:16$$

C
$$25:1$$

D
$$9:4$$

Solution

Slit width ratio $$=4:9$$, hence $$I_1 : I_2=4:9$$

$$\therefore \dfrac {a_1^2}{a_2^2}=\dfrac {4}{9}\Rightarrow \dfrac {a_1}{a_2}=\dfrac {2}{3}$$

$$\therefore \dfrac {I_{max}}{I_{min}}=\dfrac {(a_1 +a_2)^2}{(a_1 -a_2)^2}=\dfrac {25}{1}$$

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 47

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Wave Optics Test - 47

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SHARING IS CARING

Question 1

$$4400\mathop {\rm{A}}\limits^{\rm{0}} $$

$$1833\mathop {\rm{A}}\limits^{\rm{0}} $$

$$3167\mathop {\rm{A}}\limits^{\rm{0}} $$

$$7333\mathop {\rm{A}}\limits^{\rm{0}} $$

Solution

Question 2

0.1mm

1cm

0.1 cm

1m

Solution

Question 3

the image of the housefly will be reduced

there is a reduction in the intensity of the image

there is an increase in the intensity of the image

the image of the housefly will be enlarged

Solution

Question 4

$$\dfrac {\sqrt {n}}{n+1}$$

$$\dfrac {2 \sqrt {n}}{n+1}$$

$$\dfrac {\sqrt {n}}{(n+1)^{2}}$$

$$\dfrac {2 \sqrt {n}}{(n+1)^{2}}$$

Solution

Question 5

$${I_0}$$

$${I_0}$$/2

3$${I_0}$$/4

$${I_0}$$/3

Solution

Question 6

much smaller than wavelength of light.

of same order as wavelength of light

much larger than wavelength of light

none of the above

Solution

Question 7

50

100

141

400

Solution

Question 8

$$\sqrt { \lambda } \approx \cfrac { { d } }{ 4D } $$

$$\lambda \approx \cfrac { { d }^{ 2 } }{ 4D } $$

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

$$\lambda \approx \cfrac { D }{ 4 } $$

Solution

Question 9

the intensities of individual source are $$5$$ and $$4$$ units respectively

the intensities of individual source are $$4$$ and $$1$$ units respectively

the ratio of their amplitude is $$3$$

the ratio of their amplitude is $$4$$

Solution

Question 10

$$169:25$$

$$81:16$$

$$25:1$$

$$9:4$$

Solution

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