Question

A diverging meniscus lens of 1.5 refractive index has concave surface of radii 3 and 4 cm.The position of the image, if an object is placed 12 cm in front of the lens, is

• A. 7 cm
• B. -8 cm
• C. 9 cm
• D. 10 cm

Answer 1

The correct option is B -8 cm

After the refraction due to the first concave surface, a primary image is formed at a distance $v_1$. Now this image acts as virtual object for the refraction due to second concave surface and thus a secondary image is formed at a distance $v_2$ from the lens.

Using $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{({\mu }_2-{\mu }_1)}{R}$

For refraction due to first concave surface :

${\mu }_1=1$, ${\mu }_2=1.5$, $R=-3$ cm, $u=-12$ cm and $v=v_1$ .

$\therefore$ $\frac{1.5}{v_1}-\frac{1}{-12}=\frac{(1.5-1)}{-3}$

$⟹$ $v_1=-6$ cm

Thus primary image is formed at $-6$ cm in front of the lens. Now this image acts as virtual object of secondary image.

For refraction due to second concave surface :

${\mu }_1=1.5$, ${\mu }_2=1$, $R=4$ cm, $u=-6$ cm and $v=v_2$ .

$\therefore$ $\frac{1}{v_2}-\frac{1.5}{-6}=\frac{(1-1.5)}{4}$

$⟹$ $v_2=-8$ cm

Thus image is formed at $-8$ cm in front of the diverging meniscus lens.