2203 Views

in fraunhofer diffraction pattern slit width is 0.2mm and screen is at 2m away from the lens if wavelength of light used is 5000A then the distance between the first minimum on either side of the central maximum is


Vaishnavi Shirke 19th Apr, 2020
Answer (1)
Smrity 19th Apr, 2020

Hey,

We know that,

the distance between the first 2 minimum of either side of the central maximum is equal to 2 lambda D/ d

where,

it is given that,

d=0.2mm

D= 2 m

lambda=5000A

is,

= (2*5*10^3*10^-10*2)/(2*10^-1*10^-3)

=10*(10^-7/10^-4)

=10^-2 m

which is the required answer.

I hope this helps.

All the best!

Related Questions

East Point College | BBA Admi...
Apply
NBA Accredited | AICTE Approved
TAPMI MBA 2025 | Technology M...
Apply
MBA Admission Open in Technology Management and AI & Data Science | NAAC A++ | Institution of Eminence | Assured Scholarships
Amity University, Noida Law A...
Apply
700+ Campus placements at top national and global law firms, corporates, and judiciaries
East Point College | MBA Admi...
Apply
NBA Accredited | AICTE Approved
Chitkara University MBA Admis...
Apply
NAAC A+ Accredited | 100% CAMPUS RECRUITMENT
Amrita Vishwa Vidyapeetham | ...
Apply
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships | Extended Application Deadline: 30th Jan
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books