Question

Choose the correct answer from alternatives given.
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. -280.0 cm
• B. 40.0 cm
• C. 21.5 cm
• D. 13.3 cm

Answer 1

Answer: (B)

40.0 cm

According to lens marker's formula
$\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
For the first lens,
$n=1.5,R_1=+14cm,R_2=\infty$
$\therefore \frac{1}{f_1}=(1.5-1)(\frac{1}{14}-\frac{1}{\infty })=\frac{0.5}{14}$
For the second lens, $n=1.2,R_1=\infty ,R_2=-14cm$
$\therefore \frac{1}{f_2}=(1.2-1)(\frac{1}{\infty }-\frac{1}{-14})=\frac{0.2}{14}$
The focal length of the bi - convex lens is
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{0.5}{14}+\frac{0.2}{14}=\frac{0.7}{14}=\frac{1}{20}$
According to thin lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, u = -40 cm , f = 20 cm
$\therefore \frac{1}{v}-\frac{1}{-40}=\frac{1}{20}or\frac{1}{v}=\frac{1}{20}-\frac{1}{40}orv=40cm$