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Question

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface.
If the concave part is filled with water of refractive index 4/3, find the distance through which the pin should be moved so that the image of the pin again coincides with the pin?

160946_3850aa0de1334fb99421172fb9c91dc5.png

A
Δx=1.15cm,up
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B
Δx=3.15cm,down
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C
Δx=0.05cm,up
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D
Δx=1.15cm,down
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Solution

The correct option is C Δx=1.15cm,down
Method 1 : When the concave part is filled with water of refractive index 4/3, the optical arrangement is equivalent to concave mirror of focal length F such that
1F=1fw+1fg+1fm+1fg+1fw
1fw=(431)(1601)=1180
fw=180 cm
fg=60 cm (calculated earlier)
1F=1180+160+110+160+1180=26180
F=18026cm
X1=R=2F=18026×2=18013=13.85 cm
Method 2 : We use the equation
n2x2n1x1=n2n1R
For refraction at the interface 1 (air water),
4/3x21x1=4/31....(i)
The image of interface 1 is the object for the interface 2.
1.5+201x1=1.54/3+60
X1=36026=13.85
X=15.013.85=1.15 cm
Method 3 : Using lensmaker's formula and the relation
1F=1x21x1
fm=180 cm (using lensmaker's formula)
fg=60 cm (using lensmaker's formula)
1180=1x21x1 (for the water lens) ... (i)
160=1+201x1 (for the glass lens) ... (ii)
(The image by the water lens is the object for the glass lens and if the image by the glass lens is at +20, then the rays will fall normally on the mirror.)
Adding Eqs. (i) and (ii).
1180+160=1201x1
x1=1803=13.85 cm
x=15.013.85=1.15 cm.

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