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JEE Advanced 2015 Paper 1
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© examsnet.com
Question : 15
Total: 60
Two identical glass rods
S
1
and
S
2
(refractive index=1.5) have one convex end of radius of curvature 10 cm.They are planed with the curved surfaces at a distance d as a shown in the figure figure.with their axes (shown by the dashed line)aligned.when a point source of light P is placed inside rod
S
1
on its axis at a distance of 50 cm from the curved face ,the light rays emanating from it are found to br parallel t o the axis inside
S
2
.The distance d is
[JEE Adv 2015 P1]
60 cm
70 cm
80 cm
90 cm
Validate
Solution:
For 1st refraction
1
v
−
1.5
−
50
=
1
−
1.5
−
10
⇒
v
=
50
c
m
For 2nd refraction
1.5
∞
−
1
−
x
=
1.5
−
1
−
10
⇒
x
=
20
c
m
⇒
d
=
70
c
m
© examsnet.com
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