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 = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be _________.

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

Answer 1

The correct option is B $40.0cm$

The correct option is B.

Given parameters are:

Refractive indices ${\mu }_1=1.5$ and

${\mu }_2=1.2$

Radius of curvatures are $R_1=14cm$

And$R_2=\infty$

Object distance i.e. $u=-40cm$

The formula is:

$\frac{1}{f}=(\mu –1)\times (\frac{1}{R_1}-\frac{1}{R_2})$

Thus the combined focal length will be:

$\frac{1}{f}=({\mu }_1–1)\times (\frac{1}{R_1}-\frac{1}{R_2})+({\mu }_2–1)\times (\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{f}=(1.5–1)\times (\frac{1}{14}-\frac{1}{\infty })+(1.2–1)\times (\frac{1}{14}-\frac{1}{\infty })$

Therefore, solving it we get,

$f=20cm$

Now, lens formula we have,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

$\frac{1}{v}=\frac{1}{20}+\frac{1}{-40}$

$v=40cm.$