light fron a point source in air falls on a sperical glass surface (n=1.5 and radius of curvature =20 cm) . the distance of the light from the glass surface is 100cm . at what position the image is formed
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Answered by
111
use formula
n2/v - n1/u = (n2-n1)/R
so
1.5/v = (1.5-1)/20 + 1/-100
1.5/v = 5/200 - 1/100
1.5/v = 5-2/200
1.5/v = 3/200
v = 1.5x200/3 = 100.
n2/v - n1/u = (n2-n1)/R
so
1.5/v = (1.5-1)/20 + 1/-100
1.5/v = 5/200 - 1/100
1.5/v = 5-2/200
1.5/v = 3/200
v = 1.5x200/3 = 100.
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Answered by
43
Hey dear,
◆ Answer-
v = 1 m
◆ Explaination-
# Given-
For air, n1 = 1
For glass, n2 = 1.5
R = 20 cm = 0.2 m
u = -100 cm = -1 m
# Solution-
Using formula for refraction at single spherical surface -
n2/v - n1/u = (n2-n1)/R
1.5/v + 1/1 = (1.5-1)/0.2
1.5/v = 2.5 - 1
v = 1.5/1.5
v = 1 m
v = 100 cm
Image will be formed at 1 m.
Hope this helps you...
◆ Answer-
v = 1 m
◆ Explaination-
# Given-
For air, n1 = 1
For glass, n2 = 1.5
R = 20 cm = 0.2 m
u = -100 cm = -1 m
# Solution-
Using formula for refraction at single spherical surface -
n2/v - n1/u = (n2-n1)/R
1.5/v + 1/1 = (1.5-1)/0.2
1.5/v = 2.5 - 1
v = 1.5/1.5
v = 1 m
v = 100 cm
Image will be formed at 1 m.
Hope this helps you...
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