If one surface of a double convex lens has been
silvered then its effective focal length f is (take
refractive index of material as u and radius of
curvature of either surface as R)
R.
(1)
RC 3
(2)
M-2
R
R.
(3)
(4)
4u-2
2u
Answers
Answered by
1
Answer:
Given
Radius of curvature =R
focal length =f
Refractive index, μ=1.5
From lens maker's formula
f
1
=(μ−1)[
R
1
1
−
R
2
1
]
f
1
=(1.5−1)[
R
1
+
R
1
]
=0.5×
R
2
f
1
=
R
1
f=R
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