Question

A slit of width d is illuminated by a red light of wavelength 6500a°. For what value of d will 1.The first minimum fall at an angle of diffraction of 30° and 2. The first maximum fall at angle of diffraction of 30°.

Answer 1

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Answer 2

Answer:

The value of d for minimum and maximum is $1.3\mu\ m$ and $1.95\mu\ m$.

Explanation:

Given that,

Wave length $\lambda= 6500\times10^{-10}\ m$

Angle of diffraction = 30°

We know that,

Formula of first order minima

$d\sin\theta=n\lambda$

$d = \dfrac{n\lambda}{\sin\theta}$....(I)

Formula of first order maxima

$d\sin\theta=(n+\dfrac{1}{2})\lambda$

$d=\dfrac{3\lambda}{2\sin\theta}$....(II)

(I). The first minimum fall at an angle of diffraction of 30°

Using equation (I)

$d = \dfrac{n\lambda}{\sin30^{\circ}}$

$d = \dfrac{1\times6500\times10^{-10}}{\dfrac{1}{2}}$

$d = 1.3\times10^{-6}\ m$

$d = 1.3\mu\ m$

(II). The first maximum fall at an angle of diffraction of 30°

Using equation (II)

$d=\dfrac{3\times6500\times10^{-10}}{2\sin30^{\circ}}$

$d=\dfrac{3\times6500\times10^{-10}}{2\times\dfrac{1}{2}}$

$d =1.95\times10^{-6}\ m$

$d = 1.95\mu\ m$

Hence, The value of d for minimum and maximum is $1.3\mu\ m$ and $1.95\mu\ m$