Home
Class 12
PHYSICS
If intensity in YDSE is 50% of maximum a...

If intensity in YDSE is `50%` of maximum at a point. Calculate the path difference.

Text Solution

AI Generated Solution

To solve the problem of finding the path difference when the intensity in the Young's Double Slit Experiment (YDSE) is 50% of the maximum intensity, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Intensity Formula**: The intensity \( I \) at a point in YDSE can be expressed in terms of the maximum intensity \( I_0 \) and the phase difference \( \phi \): \[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) ...
Promotional Banner

Similar Questions

Explore conceptually related problems

In Young's double-slit experimetn, the intensity of light at a point on the screen, where the path difference is lambda ,is I . The intensity of light at a point where the path difference becomes lambda // 3 is

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference is lambda is K, ( lambda being the wave length of light used). The intensity at a point where the path difference is lambda"/"4 , will be

In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is lambda is K, ( lambda being the wavelength of light used). The intensity at a point where the path difference is lambda//4 will be

Waves emitted by two identical sources produces intensity of K unit at a point on screen where path difference between these waves is lamda . Calculate the intensity at that point on screen at which path difference is (lamda)/(4)

In Young's double slit experiment, the intensity of light at a point on the screen where path difference is lambda is I. If intensity at another point is I/4, then possible path differences at this point are

In Young's double-slit experiment using monochromatic light of wavelength lambda, the intensity of light at a point on the screen where path difference is lambda , is K units. What is the intensity of lgight at a point where path difference is lambda/3 .

In young's double slit experiment using monochromatic light of wavelengths lambda , the intensity of light at a point on the screen with path difference lambda is M units. The intensity of light at a point where path difference is lambda//3 is

The maximum intensity in Young's double slit experiment is I_(0) . What will be the intensity of light in front of one the slits on a screen where path difference is (lambda)/(4) ?

(a) Why are coherent sources necessary to produce a sustained interference pattern ? (b) In Young's double slit experiment using monochromatic light of wavelength lamda , the intensity of light at a point on the screen where path difference is lamda , is K units. Find out the intensity of light at a point where path difference is lamda//3 .

In Young.s double-slit experiment using monochromatic light of wavelength lambda , the intensity of light at a point on the screen where path difference is lambda , is K units. What is the intensity of light at a point where path difference is lambda"/"3 ?