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Physics Question on Optics

Two thin biconvex lenses have focal lengths f1 and f2. A third thin biconcave lens has focal length of f3. If the two biconvex lenses are in contact, the total power of the lenses is P1. If the first convex lens is in contact with the third lens, the total power is P2. If the second lens is in contact with the third lens, the total power is P3 then

A

P1=f1f2f1−f2,P2=f1f3f3−f1 and P3=f2f3f3−f2P_1=\frac{f_1f_2}{f_1-f_2},P_2=\frac{f_1f_3}{f_3-f_1}\ \text{and}\ P_3=\frac{f_2f_3}{f_3-f_2}

B

P1=f1−f2f1f2,P2=f3−f1f3+f1 and P3=f3−f2f2f3P_1=\frac{f_1-f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_3+f_1}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}

C

P1=f1−f2f1f2,P2=f3−f1f1f3 and P3=f3−f2f2f3P_1=\frac{f_1-f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}

D

P1=f1+f2f1f2,P2=f3−f1f1f3 and P3=f3−f2f2f3P_1=\frac{f_1+f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}

Answer

P1=f1+f2f1f2,P2=f3−f1f1f3 and P3=f3−f2f2f3P_1=\frac{f_1+f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}

Explanation

Solution

The correct answer is (D) : P1=f1+f2f1f2,P2=f3−f1f1f3 and P3=f3−f2f2f3P_1=\frac{f_1+f_2}{f_1f_2},P_2=\frac{f_3-f_1}{f_1f_3}\ \text{and}\ P_3=\frac{f_3-f_2}{f_2f_3}.