   # Measurements are made of the intensity distribution within the central bright fringe in a Young’s interference pattern (see Fig. 36.5). At a particular value of y , it is found that I / I max = 0.810 when 600-nm light is used. What wavelength of light should be used to reduce the relative intensity at the same location to 64.0% of the maximum intensity? ### Physics for Scientists and Enginee...

9th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305116399

#### Solutions

Chapter
Section ### Physics for Scientists and Enginee...

9th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305116399
Chapter 37, Problem 37.58AP
Textbook Problem
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## Measurements are made of the intensity distribution within the central bright fringe in a Young’s interference pattern (see Fig. 36.5). At a particular value of y, it is found that I/Imax = 0.810 when 600-nm light is used. What wavelength of light should be used to reduce the relative intensity at the same location to 64.0% of the maximum intensity?

To determine
The wavelength of light to reduce the relative intensity.

### Explanation of Solution

Given info: The ratio of the intensity to maximum intensity for first wavelength measurement of light is 0.810 and the wavelength in the first measurement is 600nm .

Formula to calculate intensity at point in a double slit interference pattern is,

I=Imaxcos2(πydλ1L)

Here,

λ1 is the wavelength in the first measurement.

d is the distance between the two slits.

L is the distance between two slits.

y is the fringe separation.

I is the relative intensity at certain point for first measurement.

Imax is the maximum intensity for first measurement.

Rearrange the above equation to get the πydL .

πydL=λ1cos1(IImax)12 (1)

Substitute 600nm for λ and 0.810 for IImax in equation (1) to find the πydL .

πydL=(600nm)cos1(0.810)12=271nm

The relative intensity is 64.0% of the maximum intensity for the second wavelength.

I2=64.0%I2maxI2I2max=0

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