Wave Optics Test

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Wave Optics Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

If one of the two slits of a Young's double-slit experiment is painted so that it transmits half the light intensity as the second slit, then

A
the fringe system will altogether disappear

B
the bright fringes will become brighter and the dark fringes will become darker

C
both dark and bright fringes will become darker

D
dark fringes will become brighter and bright fringes darker

Solution

Case 1: When slit are of equal width, if $$I_0$$ is the intensity of light
$$I_{max}$$ = $$I_0$$ + $$I_0$$ + 2$$ \sqrt{I_0 I_0} $$ = 4 $$I_0$$
$$I_{max}$$ = $$I_0$$ + $$I_0$$ - 2$$ \sqrt{I_0 I_0} $$ = 2$$I_0$$
Case 2: When one of the slit is painted over, its intensity would reduce to half then,
$$I_{new max}$$ = $$ \frac{I_0}{2}$$ + $$I_0$$ + 2$$ \sqrt{ \frac{I_0 I_0}{2}} $$ = 2.932 $$I_0$$
$$I_{new min}$$ = $$ \frac{I_0}{2}$$ + $$I_0$$ - 2$$ \sqrt{ \frac{I_0 I_0}{2}} $$ = 0.068 $$I_0$$
Therefore,
The contrast between bright and dark fringes is determined by intensity ratio.
Question 2
1 / -0

Light waves travel in vacuum along the y-axis. Which of the following may represent the wavefront?

A
$$ x= constant $$

B
$$y=constant$$

C
$$z=constant$$

D
$$x + y +z=constant$$

Solution

Since we know that velocity of light is perpendicular to the wavefront.
Hence y = constant represent the wavefront.
Question 3
1 / -0

In a Young's interference experiment, the central bright fringe can be identified due to the fact that it

A
has greater intensity than other fringes which are bright

B
is wider than the other bright fringes

C
is narrower than the other bright fringes

D
can be obtained by using white light instead of monochromatic light

Solution

In a Young's interference experiment, the central bright fringe can be identified due to the fact that bright light gives a general illumination at the central maxima. And this can be obtained only by using white light instead of monochromatic light.
Question 4
1 / -0

In YDSE, when a glass plate of refractive index $$1.5$$ and thickness $$t$$ is placed in the path of one of the interfering beams (wavelength $$\lambda$$), intensity at the position where central maximum occured previously remains unchanged. The minimum thickness of glass plate is

A
$$2\lambda$$

B
$$(2/3)\lambda$$

C
$$\lambda/3$$

D
$$\lambda$$

Solution

If after placing the plate, intensity at the position of central maxima position remains unchanged, then it means first maxima takes position of central maxima. In case of minimum thickness of plate 2, path difference created by the plate should be equal to $$\lambda$$.

i.e., $$t(\mu-1)=\lambda$$

$$t\left(\dfrac{3}{2}-1\right)=\lambda$$ $$\Rightarrow$$ $$t=2\lambda$$
Question 5
1 / -0

In a double-slit experiment, instead of taking slits of equal width, 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 increases

B
the intensities of maximum increases and the minima has zero intensity

C
the intensity of the maxima decreases and that of minima increases

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

Solution

When slits of equal width are taken, then intensity at maxima is $$4a^{2}$$ and at minima it is zero $$(I \propto w).$$
When one slit is doubled, then intensity at maxima will increase whereas intensity at minima will not be equal to zero and will be finite.
Question 6
1 / -0

If $$\beta$$ is the bandwidth, in Young's double slit experiment, the distance between the first dark hand and sixth bright band is:

A
$$5\dfrac{1}{2}\beta$$

B
$$6\beta$$

C
$$11\beta$$

D
$$5\beta$$

Solution

As we know,
Bandwidth $$(\beta)$$ is the distance between consecutive bright or drank bond.
As we know for bright band.
$$y = \dfrac{n\lambda D}{d}$$ $$n = 0, 1 , 2.....$$
for 6 bright band.
$$y_b = \dfrac{6\lambda D}{d}$$ ...(i)
for 1st dark band
$$y = \dfrac{(2n-1)}{2} \dfrac{\lambda D}{d} = \dfrac{(2\times 1 - 1)}{2} \dfrac{\lambda D}{d}$$

$$y_d = \dfrac{\lambda D}{2d}$$ ....(ii)

Now,
Distance '$$d''$$' b/w 1st dark and $$b^{th}$$ bright band is
$$d'' = b\dfrac {\lambda D}{d} - \dfrac{\lambda D}{2d}$$

$$d'' = 5\dfrac{1}{2} \dfrac{\lambda D}{d}$$

since $$\beta = \dfrac{\lambda D}{d}$$

$$d'' = 5 \dfrac{1}{2}\beta$$
Question 7
1 / -0

Out of the following statement which is not correct.

A
When unpolarised light passes through a Nicol's prism, the emergent light is elliptically polarised

B
Nicol's prism works on the principle of double refraction and total internal reflection

C
Nicol's prism can be used to produce and analyse polarised light

D
Calcite and Quartz are both doubly refracting crystals

Solution

It magnitude of light vector varies periodically during it's rotation, the tip of vector traces an ellipse and light is said to be elliptically polarised. This is not in nicol prism.
Question 8
1 / -0

A bulb of $$100\ watt$$ is hanging at a height of one meter above the centre of a circular table of diameter $$4\ m$$. If the intensity at a point on its rim is $$I_0$$, then the intensity at the centre of the table will be

A
$$I_0$$

B
$$2 \sqrt 5I_0$$

C
$$2 I_0$$

D
$$5 \sqrt 5I_0$$

Solution

The illuminanace at $$B$$
$$T_B=\dfrac{L}{I^2} ...(i)$$
and The illuminanace at $$C$$
$$I_C=\dfrac{L \cos \theta}{(\sqrt 5)^2}=\dfrac{L}{(\sqrt 5)^2} \times \dfrac{1}{\sqrt 5}$$
$$\Rightarrow I_C=\dfrac{L}{5 \sqrt 5} ...(ii)$$
From equation $$(i)$$ and $$(ii)\ I_B=5 \sqrt 5I_0$$
Question 9
1 / -0

Young's experiment establishes that

A
Light consists of waves

B
Light consists of particles

C
Light consists of neither particles nor waves

D
Light consists of both particles and waves

Solution

Young's double slit experiment establishes the phenomenon of interference that is caused due to constructive or destructive forms of wavelength of light source. This also establishes the notion of wave nature of light. Hence, his experiment establishes that light consists of waves.
So, option (A) is the correct answer.
Question 10
1 / -0

Colours of thin films result from or
On a rainy day, a small oil on water show brilliant colours. This is due to

A
Dispersion of light

B
Interference of light

C
Absorption of light

D
Scattering of light

Solution

Colour's of thin film are due to interference of light.

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Wave Optics Test - 61

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Wave Optics Test - 61

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Question 1

the fringe system will altogether disappear

the bright fringes will become brighter and the dark fringes will become darker

both dark and bright fringes will become darker

dark fringes will become brighter and bright fringes darker

Solution

Question 2

$$ x= constant $$

$$y=constant$$

$$z=constant$$

$$x + y +z=constant$$

Solution

Question 3

has greater intensity than other fringes which are bright

is wider than the other bright fringes

is narrower than the other bright fringes

can be obtained by using white light instead of monochromatic light

Solution

Question 4

$$2\lambda$$

$$(2/3)\lambda$$

$$\lambda/3$$

$$\lambda$$

Solution

Question 5

the intensities of both the maxima and the minima increases

the intensities of maximum increases and the minima has zero intensity

the intensity of the maxima decreases and that of minima increases

the intensity of maxima decreases and the minima has zero intensity

Solution

Question 6

$$5\dfrac{1}{2}\beta$$

$$6\beta$$

$$11\beta$$

$$5\beta$$

Solution

Question 7

When unpolarised light passes through a Nicol's prism, the emergent light is elliptically polarised

Nicol's prism works on the principle of double refraction and total internal reflection

Nicol's prism can be used to produce and analyse polarised light

Calcite and Quartz are both doubly refracting crystals

Solution

Question 8

$$I_0$$

$$2 \sqrt 5I_0$$

$$2 I_0$$

$$5 \sqrt 5I_0$$

Solution

Question 9

Light consists of waves

Light consists of particles

Light consists of neither particles nor waves

Light consists of both particles and waves

Solution

Question 10

Dispersion of light

Interference of light

Absorption of light

Scattering of light

Solution

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