Saved Bookmarks
| 1. |
light passes through narrow slits with a separation of 0.4mm.On a screen 1.6 away,the distance b/w 2 second order maximum 2.5mm. what is wavelength of light used |
|
Answer» Answer: ANSWER Separation is GIVEN by y= d nλD
where d=3mm=3×10 −3
D=2m λ 1 =480nm=480×10 −9 m λ 2 =600nm=600×10 −9 m 1 =n 2 =5 y 2 −y 1 =? So, y 1 = d nλ 1 D
y 1 = 3×10 −3
5×480×10 −9 ×2
y 1 =1.6×10 −3 m Also, y 2 = d nλ 2 D
y 2 = 3×10 −3
5×600×10 −9 ×2
y 2 =2×10 −3 m As y 2 >y 1
y 2 −y 1 =2×10 −3 −1.6×10 −3
=4×10 − 4m Therefore the separation on the screen between the fifth order bright fringes of the two INTERFERENCE patterns is 4×10 −4 m Explanation: |
|