The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is [2025]

1 4

2 8

3 6

4 5

Ans.
(4)

Fringe width of $λ_{1}$ = 480 nm

$β_{1} = \frac{λ_{1} D}{d} = 480 (\frac{D}{d})$

Similarly fringe width for $λ_{2}$ = 600 nm

$β_{2} = λ_{2} (\frac{D}{d}) = 600 (\frac{D}{d})$

Clearly we can see that for minimum value

$5 (β_{1}) = 4 (β_{2})$

That means $5^{th}$ maxima of $λ_{1}$ = 480 nm

coincides with $4^{th}$ maxima of $λ_{2}$ = 600 nm

Want Us to Email you About Special Offers & Updates?