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EAMCET Engineering 2005 Solved Paper
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© examsnet.com
Question : 24
Total: 160
The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is
greater than zero but less than 1.5
greater than 1.5 but less than 2.0
greater than 2.0 but less than 2.5
greater than 2.5 but less than 2.0
Validate
Solution:
Focal length of lens,
1
f
=
(
µ
−
1
)
(
1
R
1
−
1
R
2
)
For equi-convex lens,
R
1
=
+
R
,
R
2
=
−
R
∴
1
f
=
(
µ
−
1
)
(
1
R
−
1
−
R
)
1
f
=
(
µ
−
1
)
(
2
R
)
f
=
R
2
(
µ
−
1
)
f
<
R
,
so,
2
(
µ
−
1
)
<
1
i.e.
(
µ
−
1
)
<
1
2
(
µ
−
1
)
<
0.5
µ
<
1.5
Focal length of convex lens is positive. So,
µ
cannot be negative, hence
µ
should be greater than zero but less than 1.5 .
© examsnet.com
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