#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# in ydse two slits are 1mm apart and screen is placed 1m away from slits. calculate the fringe width when light of wave length 500 nm is used?ii) what should be the width of each slit in order to obtain 10maxima of the double slits pattern within central maximum of single slit pattern?

Arun
25763 Points
2 years ago

We need to obtain 10 maxima of double slit with in central maxima of single slit

For single slit, angular width of central maxima = $\frac{2\lambda }{b}$

or $\frac{y}{d}=\frac{2\lambda }{b}$or $y=\frac{2\lambda D}{b}$----------------------(1)

For interference pattern

Fringe width = $\frac{\lambda D}{d}$---------------------(2)

According to question

$\frac{2\lambda D}{b}=10.\frac{\lambda D}{d}$

$\therefore b=\frac{d}{5}=0.2mm$