# An object is placed in front of a concave mirror at a distance of 7.5 cm from it, image is formed at a distance of 30 cm from the mirror. Find the focal length of mirror when (i) Real image is formed (ii) Virtual image is formed.

** Solution:** For a concave mirror u becomes negative, $$\therefore u = -7.5 cm $$ (i) When real image is formed v is also negative $$ \therefore v = -30 cm $$

By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ -30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { -30-7.5 }{ \left( -7.5 \right) \times \left( -30 \right) } = \frac { -37.5 }{ 225 } $$ $$ \therefore f = \frac { 225 }{ -37.5 } = -6 cm $$ (ii) When virtual image is formed v is positive $$ \therefore v = 30 cm $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ 30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { 30-7.5 }{ \left( -7.5 \right) \times \left( 30 \right) } $$ $$= \frac { 22.5 }{ -225 } $$ $$ \therefore f = \frac { -225 }{ 22.5 } = -10 cm $$ Hence the focal length of mirror when real image is formed is -6 cm and when virtual image is formed at 30 cm, focal length of mirror will be -10 cm.

##### Write your Comment

Please or to post comment!

No comments yet.

MCQ on Optics (Part 1)

MCQ on Optics (Part 2)

MCQ on Optics (Part 3)

MCQ on Optics (Part 4)

MCQ on Optics (Part 5)

Write a note on Charge.