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A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15cm above the lens on its principal axis, the object coincides with its own images. On representing with another liquid, the object and the image again coincide at a distance 25cm from the lens. Calculate x, if the refractive index of the liquid is 10+x10
162339_274aaf9907224fa19629abaa0484087b.png

Solution
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Let f1 be the focal length of convex lens; radius of curvature of each curved face is R.
1f1=(m1){1R(1R)}=(μ1)2R
f1=R2(μ1)=R2(321)=R
When the space between the lens and mirror is filled by water of refractive index μ1=4/3, then the focal length of liquid concave lens f2 is
1f2=(μ11)(1R)
f2=Rμ11=R(431)=3R
The combined focal length of lenses is F1=15cm
1F1=1f1+1f2115=1R13R=313R
3R=30R=10cm
In the second case,
F2=25cm. Let μ1=μ2
1F2=1f1+1f2125=110+1f2
1f2=125110=2550
f2=503cm
f2=Rμ21μ21=Rf2=10(50/3)=35=0.6
μ2=1+0.6=1.6

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