Single-Slit Diffraction and the Effect of Refractive Index on Central Peak Width

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The discussion centers on the effects of refractive index on single-slit diffraction, specifically regarding the width of the central peak when submerged in water. It is established that the wavelength of light decreases in water due to the higher refractive index, which is 1.33 compared to air. As a result, the width of the central peak is expected to decrease when the apparatus is submerged. The calculated width of the slit in air is 170 micrometers, and adjustments to the wavelength must be made for calculations in water. Understanding the relationship between wavelength and diffraction patterns is crucial for determining the changes in peak width.
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[SOLVED] Single-Slit Diffraction

Homework Statement


You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be 17.9 millimeters.

If the entire apparatus were submerged in water, would the width of the central peak change?

a.The width would increase.
b.The width would decrease.
c.The width would not change.


Homework Equations



asin(theta) = m(lambda)



The Attempt at a Solution



Well for the first part of this question I found the width of this slit in air to be 170 micrometers.

I know this is a conceptual question, but I need help understanding exactly what happens. I know the index of refraction changes from 1 to 1.33.
 
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Hi fubag,

When the apparatus is underwater, what will change about the light?
 
lamda would change by a factor of decreasing 1.33 so the width would decrease therefore?
 
how do you calculute "width of the central peak"? Put the new lambda there.
 
ok thanks a lot!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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