Question

A thin convex lens of focal length 6cm is placed 4cm left of a thin concave lens of focal length 6cm. A point size object is placed at 24cm from convex lens. Choose the correct statements

Answer 1

Given:

• Focal length of convex lens (f₁) = 6 cm
• Focal length of concave lens(f₂)  = - 6 cm

• Object distance(u) = 24 cm [ from the convex lens]
• Distance between the convex and concave lens = 4 cm

Method:

For convex lens:

By applying the lens formula,

$\implies\sf{\dfrac{1}{f_1} =\dfrac{1}{v} - \dfrac{1}{u} }$

$\implies\sf{\dfrac{1}{6} =\dfrac{1}{v} - \dfrac{1}{-24} }$

$\implies\sf{\dfrac{1}{6} =\dfrac{1}{v} + \dfrac{1}{24} }$

$\implies\sf{ \dfrac{1}{v}= \dfrac{1}{6} - \dfrac{1}{24} }$

$\implies\sf{ \dfrac{1}{v}= \dfrac{4-1}{24} }$

$\implies\sf{ \dfrac{1}{v}= \dfrac{3}{24} }$

$\implies\sf{ v = 8 cm }$

The image will be formed at a distance of 8 cm on the other side of the convex lens.

This image will act as an object for the concave lens.

For concave lens:

• Distance of the image formed by the convex lens = 8 cm

• Distance of the image formed by the convex lens from the concave lens = 4 cm

$\implies\sf{\dfrac{1}{f_2} =\dfrac{1}{v'} - \dfrac{1}{u'} }$

$\implies\sf{\dfrac{1}{-6} =\dfrac{1}{v'} - \dfrac{1}{4} }$

$\implies\sf{ \dfrac{1}{v'}= \dfrac{1}{-6} + \dfrac{1}{4} }$

$\implies\sf{ \dfrac{1}{v'}= \dfrac{-2 +3}{12} }$

$\implies\sf{ \dfrac{1}{v'}= \dfrac{1}{12} }$

$\implies\sf{ v '= 12 cm }$

The image is formed at a distance of 12 cm from the concave lens.

Distance of the final image from the object

= 24 + 4 + 12

= 40 cm

Hence,

The distance of the final image from the object is 40 cm.

The nature of the final image is virtual.

Statements A and D are correct.