PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 31, Problem 38P

(a)

To determine

The angle of incidence.

(a)

Expert Solution
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Answer to Problem 38P

The angle of incidence is sin1(n1 1 1 n 2 ) .

Explanation of Solution

Formula used:

The expression for b can be given as,

  b=2dtanθrcosθi .

Here, b is the separation between rays, d is the thickness of plate θi is the angle of incidence and θr is the angle of refraction.

The expression for Snell’s law is given as,

  sinθi=nsinθr …… (1)

Calculation:

The differentiation of b can be calculated as,

  b=2dtanθrcosθidbdθi=2d(tanθrsinθi+ sec2θr cos2θi d θ r d θ i ) …… (2)

Differentiate equation (1).

  sinθi=nsinθrdθrdθi=cosθincosθr …… (3)

From equation (1), (2) and (3)

  dbdθi=2dn3cos3θr(sin4θi2n2sin2θi+n2)

The value of θi for maximum value of b can be calculated as,

  dbdθi=0sin4θi2n2sin2θi+n2=0θi=sin1(n 1 1 1 n 2 )

Conclusion:

Therefore, the angle of incidence is, sin1(n1 1 1 n 2 ) .

(b)

To determine

The angle of incidence.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

The angle of incidence is 48.5°

Explanation of Solution

Given:

The index of refraction of glass is n=1.60 .

Formula used:

The expression for the angle of incidence is,

  θi=sin1(n1 1 1 n 2 )

Calculation:

The angle of incidence can be calculated as,

  θi=sin1(n 1 1 1 n 2 )=sin1(1.6× 1 1 1 ( 1.6 ) 2 )=48.5°

Conclusion:

Therefore, the angle of incidence is 48.5°

(c)

To determine

The separation of the two beam.

(c)

Expert Solution
Check Mark

Answer to Problem 38P

The separation of the two beam is 2.81cm .

Explanation of Solution

Given:

The thickness of the glass plate is d=4cm

Formula used:

The expression for angle of refraction can be given as,

  θr=sin1(n1n2sinθi)

The expression for b can be given as,

  b=2dtanθrcosθi

Calculation:

The angle of refraction can be calculated as,

  θr=sin1( n 1 n 2 sinθi)=sin1(1 1.6sin( 48.5°))=27.9°

The separation of two beam can be calculated as,

  b=2dtanθrcosθi=2×(4cm)tan(27.9°)cos(48.5°)=2.81cm

Conclusion:

Therefore, the separation of the two beam is 2.81cm .

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Light in water is incident on glass at an angle 30o with the normal. If the refractive indices of water and glassare 1.33 and 1.52 respectively, what is the angle of refraction in the glass?
(a) A narrow beam of light is incident on the left side of the prism shown in the figure below. The prism is a right triangle, with two of its angles measuring 45°. The transmitted beam that exits the hypotenuse of the prism makes an angle of 8 = 22.2° with the direction of the incident beam. What is the index of refraction of the prism? (Round your answer to at least two decimal places.) 7 45.0° Ⓡ (b) What If? In part (a), we assumed the beam was monochromatic. Consider instead the case where the beam was composed of white light. Because the index of refraction differs for different wavelengths, the white light would be dispersed into constituent colors. Assume the index of refraction for blue wavelengths is 1.01n and for red wavelengths it is 0.99n, where n is the index of refraction found in part (a). What is the angular spread (in degrees) between red and blue light exiting the prism?

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