Question 60: Ray Optics

Question

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

Accepted Answer

First refraction through lens
Here, u = $$-$$30 cm, f = +15 cm
Using $${1 \over v} - {1 \over u} = {1 \over f}$$
$${1 \over v} - {1 \over { - 30}} = {1 \over { + 15}}$$ or $${1 \over v} = {1 \over {15}} - {1 \over {30}}$$
v = +30 cm
$$	herefore$$ Image I
1
is real and formed at a distance of 30 cm on the right side of the lens.
Distance of image I
1
from the mirror is
30 cm $$-$$ 10 cm = 20 cm
The image I
1
acts as an virtual object for the mirror. The mirror forms an image I
2
at a distance of 20 cm in front of it.
The image I
2
acts as an object for the lens.
Second refraction through lens
Here, u = +10 cm, f = +15 cm
$${1 \over v} - {1 \over { + 10}} = {1 \over { + 15}}$$ or $${1 \over v} = {1 \over {15}} + {1 \over {10}}$$ $$\Rightarrow$$ v = +6 cm
The final image I
3
is real and at a distance of 16 cm from the mirror.

Answer

virtual and at a distance of 16 cm from the mirror.

Answer

real and at a distance of 16 cm from the mirror.

Answer

virtual and at a distance of 20 cm from the mirror.

Answer

real and at a distance of 20 cm from the mirror.

Ray Optics

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

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