Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
418 views
in Physics by (87.1k points)
closed by
The distance between two coherent sources is 0.1 mm. The fringe-width on a screen 1.2 m away from the source is 6.0 mm. The wavelength of light used is
A. `4000 Å`
B. `5000 Å`
C. `6000 Å`
D. `7200 Å`

1 Answer

0 votes
by (84.5k points)
selected by
 
Best answer
Correct Answer - B
`d=0.1 mm=10^(-4)m, D=1.2 m`,
`beta=6 mm=6xx10^(-3)m`
`beta=(Dlambda)/d`
`6xx10^(-3)=(1.2xxlambda)/10^(-4) implies lambda=5xx10^(-7)m=5000 Å`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...