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.

Sunil Bhardwaj

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.

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