Thanks for the help 4 You are correct Your receipt no. is 159-8909Previous Tries

Question

Thanks for the help

4 You are correct Your receipt no. is 159-8909Previous Tries The maximum intensity observed on the screen is 1.18 Wm2. If one of the slits is covered up, what would the intensity on the screen at the center of the pattern become? 0.664 W/m 2 You are correct. Your receipt no. is 159-80410 Previous Tries The wavelength of the laser light is 0.635 /n, and each tick mark on the intensity pattern corresponds to a distance of 1.87 cm. What is the distance to the screen? 3.13m Submit Answer Incorrect. Tries 1/5 Previous Tries

Explanation / Answer

Given

slit width d = 26*10^-6 m

here 4 slits are illuminated ,

wavelength of the laser is lambda = 0.635*10^-6 m ,

dx = 1.87 cm

from the given data the first minimum will occur at a distance y

y = 5*dx = 5*0.0187 m = 0.0935 m

from the geometry Tan theta = y/R

where R is the separation from slit to screen

for smaller angles we write tan theta = sin theta

tan theta = y/R

using equatio d sin theta = m*lambda

26*10^-6 (0.0935/R) = 1*0.635*10^-6

solving for R

R = 3.828 m

the distance to the screen is R = 3.828 m

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