Question

illuminated by a laser with a wavelength of 534 nm. The interference pattern on a screen located x = 5.30 m away shows that the fourth-order bright fringe is located y = 7.70 cm away from the central bright fringe. Calculate the distance between the two slits.

The screen is now moved 2.1 m further away. What is the new distance between the central and the fourth-order bright fringe?

Answer 1

wavelength = 534 nm= 534 x10 ^-9 m

Distance of the screen from the slits D = 5.30 m

Distance of the fourth-order bright fringe from the central bright fringe y = 7.7 cm = 7.7 x10 ^-2 m

The distance between the two slits d = ?

Condition for fourth order maximum is d sin = 4

For small angles sin = tan = y/D

So, dy/D = 4

d = 4D / y

= 4(534x10 ^-9)(5.3) / (7.7x10 ^-2 )

= 1.47 x10 ^-4 m

(b). Distance D ' = D + 2.1 m

= 5.3 m + 2.1 m

= 7.4 m

We know dy '/D ' = 4

From this y ' = 4D ' / d

= 4(534x10 ^-9 )(7.4)/(1.47 x10 ^-4 )

= 0.1075 m

= 10.75 cm

   

The screen is now moved 2.1 m further away. What is the new distance between the central and the fourth-order bright fringe