When an incident light ray reflects off a boundarywith a medium having a higher index ofrefraction, the phenomenon is called:O hard reflectionO soft reflectionO soft refractionO hard refractionProblem 3 i.For the parts of the below question, fill in theempty boxes with your answer (YOUR ANSWERMUST BE ONLY A NUMBER; DO NOT WRITEUNITS; DO NOT WRITE LETTERS).A thin film of a material (nfm = 1.35) issuspended within a wire frame in air, as shown inthe figure. The arrangement is illuminated by4.8 x 10-7 m light that falls at an angle of 25" onthe surface of the film. A large region of the filmappears bright.nairb)nfilmnaira) [The sine of the angle of refraction0, at the air-film interface is given by:1.35 sin(25)sin(25°)1.359,-c)in the film.Calculate 0, in degrees.select one:- Calculate the wavelength A, of lightd) EThe minimum thickness d of the filmis obtained for a value of m such that:Om = 0Om = 1Om = 2Om = 3d=Td1Osin(25%)e) iAssume 8, = 33° and A/4.5 x 10-7 m. Calculate d.=

When a light ray reflects from a medium having a higher index of reflection it is known as hard…

Answered: When an incident light ray reflects off…

When an incident light ray reflects off a boundary with a medium having a higher index of refraction, the phenomenon is called:

University Physics Volume 3

17th Edition

ISBN:9781938168185

Author:William Moebs, Jeff Sanny

Publisher:William Moebs, Jeff Sanny

Chapter3: Interference

Section: Chapter Questions

Problem 47P: To save money on making military aircraft invisible to radar, an inventor decides to coat them with...

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Question

When an incident light ray reflects off a boundary
with a medium having a higher index of
refraction, the phenomenon is called:
O hard reflection
O soft reflection
O soft refraction
O hard refraction
Problem 3 i.
For the parts of the below question, fill in the
empty boxes with your answer (YOUR ANSWER
MUST BE ONLY A NUMBER; DO NOT WRITE
UNITS; DO NOT WRITE LETTERS).
A thin film of a material (nfm = 1.35) is
suspended within a wire frame in air, as shown in
the figure. The arrangement is illuminated by
4.8 x 10-7 m light that falls at an angle of 25" on
the surface of the film. A large region of the film
appears bright.
nair
b)
nfilm
nair
a) [
The sine of the angle of refraction
0, at the air-film interface is given by:
1.35 sin(25)
sin(25°)
1.35
9,-
c)
in the film.
Calculate 0, in degrees.
select one:
- Calculate the wavelength A, of light
d) E
The minimum thickness d of the film
is obtained for a value of m such that:
Om = 0
Om = 1
Om = 2
Om = 3
d=
T
d
1
Osin(25%)
e) i
Assume 8, = 33° and A/
4.5 x 10-7 m. Calculate d.
=

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