In Young’s double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used.  A person standing near the slits is looking at the fringe pattern.  When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits.  If the angular resolution of the eye is     the value of d0 is close to : Option 1)  1 mm Option 2) 2 mm Option 3) 4 mm Option 4)  3 mm

As we learnt in

Fringe Width -

$\beta = \frac{\lambda D}{d}$

- wherein

$\beta = y_{n+1}-y_{n}$

$y_{n+1}=$ Distance of$\left ( n+1 \right )^{th}$

Maxima$= \left ( n+1 \right )\frac{\lambda D}{d}$

$y_{n}=$Distance of $n^{th}$

maxima $= \frac{n\lambda D}{d}$

For a particular distance d0 between the slits, the eye is not able to resolve two consecutive bright fringes.

Now $\theta=\frac{\beta}{D}$  but  $\beta=\frac{\lambda}{d_{0}}$

$\therefore\ \; \theta=\frac{\lambda d_{0}}{d_{0}}=\frac{600\times 10^{-9}m}{\frac{1}{60}\times \frac{\pi}{180}rad}=2.06\times10^{-3}m \simeq2mm$

Correct option is 2.

Option 1)

1 mm

This is an incorrect option.

Option 2)

2 mm

This is the correct option.

Option 3)

4 mm

This is an incorrect option.

Option 4)

3 mm

This is an incorrect option.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-