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CGPET 2021 Solved Paper

Section: Physics

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Question : 2 of 150

Marks: +1, -0

In Young's double slit experiment, the ratio of amplitude of light coming from two slits is

2:3

I_0

be the maximum intensity, then the resultant intensity

I

when they interfere at path difference

\; \frac{\lambda}{3}

(\lambda

is the wavelength of light used)

$\; \frac{7}{25} I$
$\; \frac{9}{25} I$
$\; \frac{5}{7} I_0$
$\; \frac{3}{25} I$

Solution:

Ratio of amplitude,

a_1 : a_2 = 2 : 3

⇒

\frac{a_1}{a_2} = \frac{2}{3}

∴ Ratio of intensities,

\frac{I_1}{I_2} = \frac{a^2_1}{a^2_2} = \frac{4}{9}

Let

I_1 = 4x \text{ and } I_2 = 9x

Maximum intensity is given as

I_0 = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{4x} + \sqrt{9x})^2

= 25x

........(i)

When path difference

= \frac{\lambda}{3}

, then

phase difference

= \frac{2\pi}{\lambda} \times

path difference

= \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} = 120^{\circ}

∴

I = I_1 + I_2 + 2 \sqrt{I_1 I_2 \cos 120^{\circ}}

= 4x + 9x + 2 \sqrt{4x \times 9x \cos 120^{\circ}}

= 13x - 6x

⇒

I = 7x

........(ii)

Hence,

I/I_0 = \frac{7x}{25x} \Rightarrow I = \frac{7}{25} . I_0

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