The Nature of Light

  • #1
a) A light with a wavelength of 500 nm illuminates two narrow slits that are 0.10 mm apart. If the viewing screen is 1.20 m from the slits calculate the distance between each maximum.

b)how far would the fourth maximum be from the central maximum?



a) Given: lambda = 500 x 10e-9 m
d= 0.1 x 10e-3 m
L= 1.2 m

delta x = lambda x L/d
=(500 x 10e-9 m) (1.2 m)/(0.1 x 10 e-3 m)
=0.0006 m
=6.0 x 10 e-3

Therefore, the distance between each maximum is 6.0 x 10 e-3 m

b) Since the distance between each maximum is 6.0 x 10 e-3 m the fourth maximum would be 4 x (6.0 x 10 e-3 m)

Therefore, 4 x (6.0 x 10 e-3 m) = 2.4 x 10 e-2

Therefore, the fourth maximum would be 2.4 x 10 e-2 m away from the central maximum.


Does this look like I'm doing this question right to anybody?
 

Answers and Replies

  • #2
hage567
Homework Helper
1,509
2
Yes, you've got it right.
 

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