Question

An object of 2cms height is placed at 30cms infront of a convex mirror of focal length 15 cm.Then find Image distance and height of the image

Answer 1

Given :

• Height of the object $( \sf{h_{o}} )$ = 2cm
• Object distance $( \sf{u} )$ = 30cm
• Focal length of the mirror $( \sf{f} )$ = 15cm

To Find :

• Image distance $( \sf{v} )$
• Height of the image $( \sf{h_{i}} )$

By Applying New Cartesian Sign Conventions, we have –

• $( \sf{h_{o}} )$ = 2cm
• $( \sf{u} )$ = - 30cm
• $( \sf{f} )$ = 15cm

★ Image distance $( \sf{v} )$ = ?

From mirror formula :

$\underline{\boxed{ \bf{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }}}$

⪼ $\sf{ \dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{ - 30} }$

⪼ $\sf{ \dfrac{1}{15} + \dfrac{1}{30} = \dfrac{1}{ v} }$

⪼ $\sf{ \dfrac{2 + 1}{30}= \dfrac{1}{ v} }$

⪼ $\sf{\cancel{ \dfrac{3}{30}}= \dfrac{1}{ v} }$

⪼ $\sf{\dfrac{1}{10}= \dfrac{1}{ v} }$

⪼ $\sf{v = 10}$

• Hence, the image distance is 10 cm.

★ Height of the image $( \sf{h_{i}} )$ = ?

From magnification formula :

$\underline{\boxed{ \bf{ \dfrac{h_{i}}{h_{o}} = - \dfrac{v}{u} }}}$

⪼ $\sf{ \dfrac{h_{i}}{2} = \cancel{-} \dfrac{10}{ \cancel{-} \: 30} }$

⪼ $\sf{ \dfrac{h_{i}}{2} = \dfrac{10}{30} }$

⪼ $\sf{30 \: h_{i} = 2 \times 10}$

⪼ $\sf{\: h_{i} = \dfrac{20}{30}}$

⪼ $\sf{\: h_{i} = \dfrac{2}{3} }$

• Hence, the height of the image is 2/3 cm.