The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4 then what is the value of x

(Assume that the field strength varies according to the slit width.) [2025]

Question

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4 then what is the value of x

(Assume that the field strength varies according to the slit width.) [2025]

Smart Achievers · Accepted Answer

(3)

${(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = \frac{9}{4}$

$\Rightarrow \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{3}{2}$

$\frac{(x + 1) d}{(x - 1) d} = \frac{3}{2}$

$\Rightarrow 3 x - 3 = 2 x + 2 \Rightarrow x = 5$

Solution

1 2

2 3

3 5

4 4

Ans.
(3)

${(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = \frac{9}{4}$

$\Rightarrow \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{3}{2}$

$\frac{(x + 1) d}{(x - 1) d} = \frac{3}{2}$

$\Rightarrow 3 x - 3 = 2 x + 2 \Rightarrow x = 5$

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Solution

1 2

2 3

3 5

4 4

Ans. (3) (I1+I2I1–I2)2=94 ⇒ I1+I2I1–I2=32 (x+1)d(x–1)d=32 ⇒ 3x–3=2x+2 ⇒ x=5

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Ans.
(3)

${(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = \frac{9}{4}$

$\Rightarrow \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{3}{2}$

$\frac{(x + 1) d}{(x - 1) d} = \frac{3}{2}$

$\Rightarrow 3 x - 3 = 2 x + 2 \Rightarrow x = 5$