A mercury lamp emits several rays of light. These rays are parallel to each other and...
A mercury lamp emits several rays of light. These rays are parallel to each other and have a wave length of 546nm. The rays pass through a lens covered by a material with a slit. The focal length of the lens is 60cm. The distance between the first maximum and the first minimum is 10.2 nm. What is the width of the slit in the covering material?
Homework Answers
wavelength =546 nm=546810^-9 m
focal length f=60 cm=60*10^-2 m
distance b/w first maximum to first minium a=10.2 nm=10.2*10^-9 m
basing on the concept of light
the width of the slit W=2f(wavelength)/a
=2*60*10^-2*546*10^-9/10.2*10^-9
=64.24 m
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