A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays, to cross the slab. Speed of light in vacuum - 3 x 108 m/s.
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7
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Dear student,
μ=sinisinr=sin45sin30=12√1/2=2‾√=1.41μ=cvv=1.41×3×108=4.23×108m/st = d/v=1 cm/423×108 cm/st=0.0023×10−8
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I hope you help !!!
Dear student,
μ=sinisinr=sin45sin30=12√1/2=2‾√=1.41μ=cvv=1.41×3×108=4.23×108m/st = d/v=1 cm/423×108 cm/st=0.0023×10−8
Regards
Answered by
20
Given :
Angle of incidence= i= 45°
angle of refraction =r= 30°
μ= sin i/ sin r
as we know that μ=C/V
sin 45/ sin 30 =3 x 10 ⁸ / v
1/√2/ 1/ 2= 3 x 10 ⁸ / v
v= 3 x 10 ⁸ / 2 m/s
Distance travelled by light in slab is
x= 1m / cos 30 = 2/√3
time taken = (2/√3) x[√2/ 3 x 10⁸]
=5.4 x 10 ⁻⁹ s
∴ time taken by the light rays, to cross the slab is 5.4 x 10 ⁻⁹ sa
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