Wave Optics Test

Result

Wave Optics Test - 56

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Question 1
1 / -0

The wavefront of a light beam is given by the equation $$x + 2 y + 3 z = c$$ , (where c is arbitrary constant) then what is the angle made by the direction of light with the y-axis?

A
$$\cos ^ { - 2 } \frac { 2 } { \sqrt { 14 } }$$

B
$$\cos ^ { - 1 } \frac { 2 } { \sqrt { 14 } }$$

C
$$\cos ^ { - 3 } \frac { 2 } { \sqrt { 14 } }$$

D
$$\cos ^ { - 4 } \frac { 2 } { \sqrt { 14 } }$$

Solution

The wave front equations must be of form $$ax+by+cz=1$$ where $${a^2} + {b^2} + {c^2} = 1$$
$$($$ $$a,b,c$$ are cosines of angles made with $$x,yx,z$$ axes respectively $$)$$
applying this to the given equation$$:$$
$${1^2} + {2^2} + {3^2} = {c^2}$$
$${c^2} = 14$$
$$c = \sqrt {14} $$
$${\cos ^{ - 1}}\dfrac{2}{{\sqrt {14} }}$$
angle made with $$y-$$ axis is $${\cos ^{ - 1}}\dfrac{2}{{\sqrt {14} }}$$
Hence,
option $$(B)$$ is correct answer.
Question 2
1 / -0

A light wave has a frequency of $$5\times { 10 }^{ 14 }Hz$$ and waves length of $$4\times { 10 }^{ -7 }m$$ in a medium. The refractive index of the medium is

A
1.33

B
1.5

C
1.25

D
1.75

Solution
Question 3
1 / -0

The intensity of sound reduces by 20 $$ \% $$ on passing through a glass slab. If sound of intensity 1 is made to cross through two such slabs, then the intensity of emergent sound will be

A
$$36 \%$$

B
$$64\%$$

C
$$40\%$$

D
$$80\%$$

Solution

Let the intensity of sound be $$100$$ after passing through a slab,
percentage reduce $$=20$$%amount of intensity reduced$$=20$$% of $$00=20$$
remaining intensity $$=100-20=80$$
percentage reduction $$=20$$%
amount of intensity reduced $$=20$$% of $$80=16$$
remaining intensity $$=80-16=64$$
total amount of intensity reduced $$=100-64=36$$
reduction percentage $$ = \dfrac{{36}}{{100}} \times 100 = 36$$%
Question 4
1 / -0

The figure shows a double slit experiment where $$ P $$ and $$ Q $$ are the slits. The path lengths $$ P X $$ and $$ Q X $$ are $$ n \lambda $$ and $$ (n+2) \lambda $$ respectively, where $$ n $$ is a whole number and $$ \lambda $$ is the wavelength. Taking the central fringe as zero, what is formed at $$ X $$

A
First bright

B
First dark

C
Second bright

D
Second dark

Solution

For brightness path difference $$=n \lambda = 2\lambda$$
So second is bright.
Question 5
1 / -0

Which one of the following phenomena is not explained by Huygens construction of wave front ?

A
$$Refraction$$

B
$$Reflection$$

C
$$Diffraction$$

D
$$Origin$$ $$of$$ $$spectra$$

Solution
Question 6
1 / -0

In a young double slit experiment, 12 fringe are observed to be formed in a certain segment of the screen when light of wavelength 600 m is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

A
12

B
18

C
24

D
30

Solution
Question 7
1 / -0

While moving from deep water to shallow water.

A
Frequency of the water waves decrease

B
Frequency of the water waves increase

C
Frequency of the water waves stays the same

D
None of the above

Solution

frequency remains same while speed and wavelength changes
Water waves travel fastest when the medium is the deepest. Thus, if water waves are passing from deep water into shallow water, they will slow down. So as water waves are transmitted from deep water into shallow water, the speed decreases, the wavelength decreases, and the direction changes.
Question 8
1 / -0

Light of wavelength 6000 is incident bria single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the slit.

A
0.15 mm

B
0.1 mm

C
0.3 mm

D
0.2 mm

Solution
Question 9
1 / -0

In a YDSE both slits produce equal intensities on the screen. A 100% transparent thin film is placed in front of one of the slits. Now the intensity of the geometrical centre of system on the screen becomes 75% of the previous intensity. The wavelength of the light is 6000A and $${ \mu }_{ glass }=1.5.$$ The thickness of the film cannot be :

A
0.2 $${ \mu }m$$

B
1.0 $${ \mu }m$$

C
1.4 $${ \mu }m$$

D
1.6 $${ \mu }m$$

Solution
Question 10
1 / -0

If a ray of light is incident on dielectric surface such that the reflected ray is maximum polarized , then the angle between incident ray and normal to the surface is

A
$$tan^{-1} (n)$$

B
$$tan^{-1} (1/n)$$

C
$$sin^{-1} (n)$$

D
$$sin^{-1} (1/n)$$

Solution