Question

A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such taht its image is real and magnified and one end touches the rod. Its magnification will be –

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

End A of the rod acts as an object for mirror and A' will be its image so

|u| = (2f – l)

= (20 – 5) = 15 cm

By mirror formula

$\frac { 1 } { \mathrm { f } }$= $\frac { 1 } { - 15 }$ +

̃ $\frac { 1 } { 30 }$ ̃ V = – 30 cm

Now m = $\frac { ( 30 - 20 ) } { 5 }$ = 2

End A of the rod acts as an object for mirror and A' will be its image so

|u| = (2f – l)

= (20 – 5) = 15 cm

By mirror formula

$\frac { 1 } { \mathrm { f } }$= $\frac { 1 } { - 15 }$ +

̃ $\frac { 1 } { 30 }$ ̃ V = – 30 cm

Now m = $\frac { ( 30 - 20 ) } { 5 }$ = 2