Question

1. A concave mirror of focal length 20 cm forms a real image at a distance of 40 cm Find the position of the object O -40 cm O 40 cm O -20 cm O 20 cm​

Answer 1

Provided that:

• A concave mirror of focal length 20 cm forms a real image at a distance of 40 cm. Find the position of the object.

According to sign convention,

• Focal length = -20 cm
• Image distance = -40 cm

To calculate:

• Position of object

Solution:

• Position of object = C
• Object distance = -40 cm

Using concept:

• Mirror formula

Using formula:

• Mirror formula:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}}}}}$

Where, v denotes image distance, f denotes focal length and, u denotes object distance.

Knowledge required:

• If the magnification produced by a spherical mirror is in negative then the mirror is always “Concave Mirror.”

• If the magnification produced by a spherical mirror is in positive then the mirror is always “Convex Mirror.”

• If magnification is negative in a concave mirror then it's nature is “Real and Inverted” always.

• If magnification is positive in a concave mirror then it's nature is “Virtual and Erect” always.

• If in the ± magnification, magnitude > 1 then the image formed is “Enlarged”.

• If in the ± magnification, magnitude < 1 then the image formed is “Diminished”.

• If in the ± magnification, magnitude = 1 then the image formed is “Same sized”.

• If the focal length is positive then the mirror is “Convex Mirror.”

• If the focal length is negative then the mirror is “Concave Mirror.”

Required solution:

$:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{-40} + \dfrac{1}{u} = \dfrac{1}{-20} \\ \\ :\implies \sf \dfrac{1}{u} = \dfrac{1}{-20} + \dfrac{1}{40} \\ \\ :\implies \sf \dfrac{1}{u} = \dfrac{2 \times -1 + 1 \times 1}{40} \\ \\ :\implies \sf \dfrac{1}{u} = \dfrac{-2 + 1}{40} \\ \\ :\implies \sf \dfrac{1}{u} = \dfrac{-1}{40} \\ \\ :\implies \sf 1 \times 40 = u \times -1 \\ \\ :\implies \sf 40 = -u \\ \\ :\implies \sf u = -40 \\ \\ :\implies \sf Image \: distance = -40 \: cm$

The object is placed on the centre of curvature.