Question

The magnitudes of power of a biconvex lens (refractive index \(1.5\)) and that of a plano-convex lens (refractive index \(1.7\)) are same. If the curvature of the plano-convex lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of radii of curvature of the front and back surfaces of the biconvex lens is:

• A. \(5:2\)
• B. \(5:12\)
• C. \(12:5\)
• D. \(2:5\)

Hint:

In lens problems, always apply the lens-maker formula carefully and watch the sign convention for radii.

Answer 1

The Correct Option is D

Concept: The power \( P \) of a thin lens in air is given by the lens-maker’s formula: \[ P=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] For a plano-convex lens, one surface is plane, so its radius of curvature is infinite.
Step 1: Power of the plano-convex lens Let the curved surface radius be \( R \). Refractive index \( n=1.7 \). \[ P_p=(1.7-1)\left(\frac{1}{R}\right)=\frac{0.7}{R} \]
Step 2: Power of the biconvex lens Let the front and back radii be \( R_1 \) and \( R_2 \). Refractive index \( n=1.5 \). \[ P_b=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) =\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]
Step 3: Use the given condition Magnitudes of powers are equal and the curvature of the plano-convex lens matches the back surface of the biconvex lens: \[ R=R_2 \] \[ \left|\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\right| =\frac{0.7}{R_2} \]
Step 4: Solve for the ratio \[ \frac{1}{R_2}-\frac{1}{R_1}=\frac{1.4}{R_2} \] \[ -\frac{1}{R_1}=\frac{0.4}{R_2} \Rightarrow R_1=\frac{5}{2}R_2 \] Thus, \[ R_1:R_2=2:5 \]