Question

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surfaces are of the same radius of curvature $R=14cm$.For this bi-convex lens, for an object distance of $40cm$, the image distance will be

• A. $-280.0cm$
• B. $40.0cm$
• C. $21.5cm$
• D. $13.3cm$

Answer 1

Answer: (B)

Using the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$
or $\frac{1}{v}=\frac{1}{u}+\frac{1}{f_1}+\frac{1}{f_2}$
$15mm=\frac{1}{u}+(n_1-1)(\frac{1}{R_1}-\frac{1}{R_2})+(n_2-1)(\frac{1}{R_1^{\prime }}-\frac{1}{R_2^{\prime }})$
Substituting the values, we get
$\frac{1}{v}=\frac{1}{-40}+(1.5-1)(\frac{1}{14}-\frac{1}{\infty })+(1.2-1)(\frac{1}{\infty }-\frac{1}{14})$
Solving this equations, we get
$v=+40cm$