Wave Optics Test

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Wave Optics Test - 57

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Question 1
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A source S is kept directly behind the slit $$S$$ , in a double-slit apparatus. Find the phase difference at a point $$O$$which is equidistant from $$s 1 8$$ s2 What will be the phase difference at $$\mathrm { P }$$ if a liquid of refraction index $$\mu$$ is filled. (wavelength of light in air is $$/$$ due to the source)$$( \lambda < < d , d < < D , \langle > > d )$$

A
between the screen and the slits.

B
between the slits & the source S. In this case find the minimum distance between the points on the screen where the intensity is half the maximum
intensity on the screen.

C
Beyond the slits.

D
None of these

Solution

A source S is kept directly behind the slit S , in a double-slit apparatus. Find the phase difference at a point Owhich is equidistant from s18 s2 What will be the phase difference at P if a liquid of refraction index μ is filled between the screen and the slits.
so the correct option is A.
Question 2
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Angle of polarization depends on

A
incident angle

B
angle of refraction

C
type of medium

D
the angle of reflection.

Solution
Question 3
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Select the incorrect statement

A
The frequency of a wave depends on its source and not on the medium

B
Only a transverse wave can be polarised

C
For the propagation of mechanical wave the medium must be least elastic

D
The mechanical waves are always longitudinal as liquids and gases cannot sustain shear

Solution
Question 4
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Coherent sources can be obtained:

A
only by division of wave front

B
only by division of amplitude

C
both by division of amplitude and wave front

D
none of the above

Solution
Question 5
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The ratio of the maximum and minimum intensities in the interference pattern is pattern produced by two coherent sources of light is 9:1.the intensity of used light?

A
$$4 : 1$$

B
$$25 : 1$$

C
$$36 : 1$$

D
$$49 : 1$$

Solution

$$\begin{array}{l} \dfrac { { { I_{ \max } } } }{ { { I_{ \min } } } } =\dfrac { { { { \left( { { a_{ 1 } }+{ a_{ 2 } } } \right) }^{ 2 } } } }{ { { { \left( { { a_{ 1 } }-{ a_{ 2 } } } \right) }^{ 2 } } } } \\ \dfrac { 9 }{ 1 } =\dfrac { { { { \left( { { a_{ 1 } }+{ a_{ 2 } } } \right) }^{ 2 } } } }{ { { { \left( { { a_{ 1 } }-{ a_{ 2 } } } \right) }^{ 2 } } } } \\ \dfrac { { { a_{ 1 } } } }{ { { a_{ 2 } } } } =\dfrac { 2 }{ 1 } \\ \dfrac { { a_{ 1 }^{ 2 } } }{ { a_{ 2 }^{ 2 } } } =\dfrac { { { l_{ 1 } } } }{ { { l_{ 2 } } } } =\dfrac { 4 }{ 1 } \end{array}$$
Hence, Option $$A$$ is correct.
Question 6
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In YDSE, the spacing between the slits is d and wavelength of light used is 6000 A. If the angular width of a fringle formed on a distant screen is $$1^o$$, then value of d is

A
1 mm

B
0.05 mm

C
0.03 mm

D
0.01 mm
Question 7
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In a YDSE bichromatic light of wavelength $$400nm$$ and $$560nm$$ are used. The distance between the slits is $$0.1mm$$ and the distance between the plane of the slits and the screen is $$1m$$. The minimum distance between two successive regions of complete darkness is

A
$$4mm$$

B
$$5.6mm$$

C
$$14mm$$

D
$$28mm$$

Solution

$$\begin{array}{l} Let\, \, nth\, \, \min ima\, \, of\, \, 400nm\, \, coincides\, \, with\, \, mth\, \, \min ima\, \, of\, \, 560nm \\ then,\, \, \left( { 2n-1 } \right) 400=\left( { 2m-1 } \right) 560 \\ \Rightarrow \frac { { 2n-1 } }{ { 2m-1 } } =\frac { 7 }{ 5 } =\frac { { 14 } }{ { 10 } } =\frac { { 21 } }{ { 15 } } \\ i.e.\, \, 4th\, \, \min ima\, \, of\, \, 400nm\, \, coincides\, \, with\, \, 3rd\, \, \min ima\, \, of\, \, 560nm. \\ =\frac { { 7\left( { 1000 } \right) \left( { 400\times { { 10 }^{ -6 } } } \right) } }{ { 2\times 0.1 } } =14\, mm \\ next,\, \, 11th\, \, \min ima\, \, of\, \, 400\, nm\, \, will\, \, coincide\, \, with\, \, 8th\, \, \min ima\, \, of\, \, 560\, nm \\ location\, \, of\, \, this\, \, \min ima\, \, is \\ =\frac { { 21\left( { 1000 } \right) \left( { 400\times { { 10 }^{ -6 } } } \right) } }{ { 2\times 0.1 } } =42\, mm] \\ \therefore required\, \, dis\tan ce=28\, mm \end{array}$$
Hence,
option $$(D)$$ is correct answer.
Question 8
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In a standard YDSE setup, the fringe width on the screen is 1.5 mm. When a thin glass film is pasted in front of the upper slit, the fringe pattern shifts up. But it is seen that at a point P above central maxima where intensity was one fourth the intensity at central maxima, intensity remains the same. There were no maxima between central maxima and point P before film was introduced. What can be the thickness of the film? Take $$\mu$$ = 1.5, $$\lambda$$ = 450 nm :-

A
3 x $$10^{-7}$$m

B
5 x $$10^{-7}$$m

C
9 x $$10^{-7}$$m

D
1.4 x $$10^{-6}$$m

Solution

$$\begin{array}{l} Path\, difference\, due\, to\, slab\, should\, be\, { { int } }egral\, multiple\, of\, \, \lambda \, or\, \Delta x=n\lambda \\ \left( { \mu -1 } \right) t=n\lambda \, \, \, \, \, \, \, n=1.2.3...... \\ or\, t=\frac { { n\lambda } }{ { \mu -1 } } \\ For\, { { minimum } }\, value\, of\, t\, ,\, we\, have\, n=1 \\ \therefore t=\frac { \lambda }{ { \mu -1 } } =\frac { \lambda }{ { \left( { 1.5-1 } \right) } } =2\lambda \\ \end{array}$$
On putting the value of $$\lambda $$ we get
$$t = 1.4 \times {10^{ - 6}}$$
Hence, The option $$D$$ is the correct answer.
Question 9
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In Y.D.S.E. the fringe width is 0.2 mm. If wavelength of light is increases by 10% and separation between the slits is increases by 10% then fringe width will be:

A
$$0.20 mm$$

B
$$0.165 mm$$

C
$$0.401 mm$$

D
$$.242 mm$$

Solution
Question 10
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Particle nature and wave nature of electomagnetic waves can be shown by

A
electron has small mass, deflected by the metal sheet

B
X-ray is diffracted a, reflected by thick metal sheet

C
light is refracted and diffracted

D
light is polarised and shows photoelectric effect.

Solution

Particle nature is shown by photoelectric effect and wave nature is shown by polarized light. Hence option D is correct