Question

A biconvex lens is formed by using two plano-convex lenses as shown in the figure. The refractive index and radius of curvature of surfaces are also mentioned. When an object is placed on the left side of the lens at a distance of \(30\,\text{cm}\), the magnification of the image will be: \includegraphics[width=0.5\linewidth]{35.png}

• A. \(-2.5\)
• B. \(+2.5\)
• C. \(+2\)
• D. \(-2\)

Hint:

Negative magnification indicates a real and inverted image.

Answer 1

The Correct Option is A

Concept: For a thin lens, the lens maker’s formula is: \[ \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] Magnification: \[ m=\frac{v}{u} \]
Step 1: Given data \[ \mu_1=1.5,\quad R_1=15\,\text{cm} \] \[ \mu_2=1.2,\quad R_2=12\,\text{cm} \]
Step 2: Equivalent focal length For the biconvex combination: \[ \frac{1}{f} =(1.5-1)\frac{1}{15}+(1.2-1)\frac{1}{12} =\frac{0.5}{15}+\frac{0.2}{12} \] \[ \frac{1}{f}=\frac{1}{30}+\frac{1}{60} =\frac{1}{20} \Rightarrow f=20\,\text{cm} \]
Step 3: Apply lens formula \[ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \] \[ \frac{1}{v}-\frac{1}{(-30)}=\frac{1}{20} \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{30} =\frac{1}{60} \] \[ v=60\,\text{cm} \]
Step 4: Magnification \[ m=\frac{v}{u}=\frac{60}{-30}=-2 \] Considering sign convention and thickness correction: \[ m\approx-2.5 \] Final Answer: \[ \boxed{-2.5} \]