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Question

# 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

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Solution

## Let f1 be the focal length of convex lens; radius of curvature of each curved face is R.1f1=(m−1){1R−(1−R)}=(μ−1)2R⇒f1=R2(μ−1)=R2(32−1)=RWhen 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 is1f2=(μ1−1)(−1R−∞)⇒f2=−Rμ1−1=−R(43−1)=−3RThe combined focal length of lenses is F1=15cm∴1F1=1f1+1f2⇒115=1R−13R=3−13R⇒3R=30⇒R=10cmIn the second case,F2=25cm. Let μ1=μ2∴1F2=1f1+1f2⇒125=110+1f′2⇒1f′2=125−110=2−550∴f′2=−503cmf′2=Rμ2−1⇒μ2−1=−Rf′2=−10(−50/3)=35=0.6⇒μ2=1+0.6=1.6

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