Question 273: Waves Optics

Question

In a Fraunhoffer diffraction, light of wavelength ' $\lambda$ ' is incident on slit of width ' d '. The diffraction pattern is observed on a screen placed at a distance ' $D$ '. The linear width of central maximum is equal to two times the width of the slit, then 'D' has value

Accepted Answer

Given:
Wavelength of light = $\lambda$
Width of slit = $d$
Distance to screen = $D$
Linear width of central maximum = $2d$
Let's solve step by step:
1.
Position of first minima in single-slit Fraunhofer diffraction
The angular position of first minima is given by:
$ d \sin 	heta = \pm \lambda $
For small angles, $\sin 	heta \approx 	heta$ (in radians).
So,
$ d 	heta = \lambda \implies 	heta = \frac{\lambda}{d} $
2.
Linear width of central maximum
Central maximum is between first minima on both sides:
Angular width: $2	heta = 2\left(\frac{\lambda}{d}ight)$
Linear (on screen):
$   	ext{Linear width} = 2D 	heta = 2D \left(\frac{\lambda}{d}ight) = \frac{2D\lambda}{d}   $
3.
Given condition:
$ 	ext{Linear width of central maximum} = 2d $
So,
$ \frac{2D\lambda}{d} = 2d $
4.
Solve for $D$:
Divide both sides by 2:
$ \frac{D\lambda}{d} = d $
$ D\lambda = d^2 $
$ D = \frac{d^2}{\lambda} $
5.
Final Answer:
Option A:
$ \boxed{\frac{d^2}{\lambda}} $
This is the correct value of $D$.

Answer

$\frac{\mathrm{d}^2}{\lambda}$

Answer

$\frac{\mathrm{d}^2}{2 \lambda}$

Answer

$\frac{\mathrm{d}^2}{3 \lambda}$

Answer

$\frac{\mathrm{d}^2}{4 \lambda}$

Waves Optics

In a Fraunhoffer diffraction, light of wavelength ' \lambda ' is incident on slit of width ' d '. The diffraction pattern is observed on a screen placed at a distance ' D '. The linear width of central maximum is equal to two times the width of the slit, then 'D' has value

Choose the correct answer: