Question

An object 0.04 m high is placed at a distance of 0.8 m from a concave mirror of radius of curvature 0.4 m. Find the position , nature & size of the image, the formula used will be; *
A) 1/v +1/u = 1/f
B) 1/v =1/f -1/u
C)both 1 & 2
D) 1/v -1/u = 1/f​

Answer 1

Given:

An object 0.04 m high is placed at a distance of 0.8 m from a concave mirror of radius of curvature 0.4 m.

To find:

• Formula used
• Position of image
• Image characteristics

Calculation:

Applying Mirror Formula:

$\sf \: \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$\sf \implies \dfrac{2}{r} = \dfrac{1}{v} + \dfrac{1}{u}$

$\sf \implies \dfrac{2}{( - 40)} = \dfrac{1}{v} + \dfrac{1}{ (- 80)}$

$\sf \implies - \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{ 80}$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{ 80} - \dfrac{1}{20}$

$\sf \implies \dfrac{1}{v} = \dfrac{1 - 4}{ 80}$

$\sf \implies \dfrac{1}{v} = \dfrac{ - 3}{ 80}$

$\sf \implies \: v = - 26.67 \: cm$

• So, image distance is -26.67 cm.

$\rm \: mag. = - \dfrac{v}{u}$

$\rm \implies \: mag. = - \dfrac{ - 26.67}{ - 80}$

$\rm \implies \: mag. = - 0.33$

• So, image is real , inverted and diminished.

Also, both A) and B) are correct.

Hope It Helps.