Question

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20cm. The concave surface has a radius of curvature 60 cm. the convex side is silvered and placed on a horizontal surface.
Where should a pin be placed on the optic axis such that its image is formed at the same place?

• A. 15 cm
• B. 30 cm
• C. 40 cm
• D. 45 cm

Answer 1

The correct option is (A) 15 cm

The effective focal length F of the combination is given as $\frac{1}{F}=\frac{1}{f_g}-\frac{1}{f_m}+\frac{1}{f_g}$

because refraction takes place twice with the curved surface and one reflection takes place with the mirror.

We can write,

$-\frac{1}{F}=\frac{2}{f_g}-\frac{1}{f_m}$

We know,

$\frac{1}{f_g}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$ ------------(1)

And also

$f_m=\frac{R_1}{2}$ ---------(2)

Where $R_{1}and{R}_2$ are the radius of curvature of the lower and upper surface of the lens.

Putting $R_1=20cm,R_2=60cmn=1.5$ in equation 1

$\frac{1}{f_g}=(1.5-1)(\frac{1}{20}-\frac{1}{60})$

After solving we get

$f_g=+60cm$

From equation 2

$f_m=-10cm$

Now,

We can write,

$-\frac{1}{F}=\frac{2}{60}-\frac{1}{-10}$

After solving

$F=-7.5cm$ (concave mirror)

For the image to be formed at the same points, the object distance

u=$2F$ =−15 cm

Hence, the object should be placed at a distance of 15 cm from the lens on the optic axis.