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Question
A parallel beam of monochromatic light is incident on a glass slab at an angle of incidence 60o. Find the ratio of width of the beam in the glass to that in the air if refractive index of glass is 3/2.
Solution
Given: i = 60°, μg = 1.5,
Let dg = width of beam in glass slab,
da = width of beam in air
To find: Ratio of widths `(d_g/d_a)`
Formulae:
i. `mu_g=sini/sinr`
ii. `d_g/d_a=cosr/cosi`
Calculation: From formula (i),
sin r = sin i/μg
∴ sin r = sin60°/1.5 = 0.8660/1.5 = 0.5773
∴ r = sin-1 (0.5773) = 35°16'
From formula (ii),
`d_g/d_a=cosr/cosi=(cos35^@16')/cos60^@`
`therefore d_g/d_a=0.8164/0.5=1.633`
`therefored_g/d_a=1.633:1`
∴ Ratio of the widths of beam = 1.633 : 1
The ratio of widths of the beam in glass to that in air is 1.633 : 1.
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