CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm . What is the index of refraction of the soap solution, if it is known to be between 1.2 and 1.5 ?
 


Solution

Given: d=0.0011×103m
 For minimum reflection of light , 2 md = nl
μ=nλ2d=2nλ4d=580×109×(2n)4×11×10=58(2n)44=0.132(2n)
Given that μ has a value in between 1.2 and 1.5.
When, n = 5,
μ=0.132×10=1.32
 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image