Question

The focal length of a concave lens is 30 cm. At what distance should the object be placed from the lens so that it forms an image at 20 cm from the lens

Answer 1

AnsWer :

- 60 cm.

To FinD :

The distance of object place.

SolutioN :

Let,

• u object distance.
• v image distance.
• f focal length.

Where as,

• f = 30 cm.
• v = 20 cm.

$\rule{50}1$

☯ Sign Convention.

• f = - 30 cm.
• v = - 20 cm.

☛ We know, Lens Formula.

$\tt \dagger \: \: \: \: \: \fbox{ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}$

$\tt : \implies \dfrac{1}{ - 30} = \dfrac{1}{ - 20} - \dfrac{1}{u}$

$\tt : \implies \dfrac{1}{ - 30} + \dfrac{1}{20} = - \dfrac{1}{u}$

$\tt : \implies \dfrac{ - 2 + 3}{ 60} = - \dfrac{1}{u}$

$\tt : \implies \dfrac{ 1}{ 60} = - \dfrac{1}{u}$

$\tt : \implies u = - 60 \: cm.$

✡ Therefore, When we place object at a distance of 60 cm so that image form at distance of 20 cm.

$\rule{200}3$

More InformatioN :

★ Mirror Formula.

$\tt \dagger \: \: \: \: \: \fbox{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}$

★ Lens Formula.

$\tt \dagger \: \: \: \: \: \fbox{ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}$

★ Magnification. ( for mirror )

$\tt \dagger \: \: \: \: \: \fbox{M = \dfrac{h_i}{h_o}= \dfrac{-v}{u}}$

★ Magnification. ( for lens )

$\tt \dagger \: \: \: \: \: \fbox{M = \dfrac{h_i}{h_o}= \dfrac{v}{u}}$

Answer 2

AnswEr:-

Your Answer is that the object should be 60 cm in front of the mirror.

ExplanaTion:-

Given:-

• Focal length of concave lens (f) = -30cm. As focal length of a concave lens is always negative.

• Image distance (v) = -20cm as image by concave mirror is always formed in front of lens.

To Find:-

• The object distance (u).

Formula Used:-

Lens Formula:-

$\tt\longrightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} .$

So Here,

• v = -20 cm.
• u = ?? [To Find].
• f = -30 cm.

Therefore lets put the values in formula:-

$\tt\mapsto \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} . \\ \\ \tt\mapsto \frac{1}{-20 \: cm} - \frac{1}{u} = \frac{1}{ - 30 \: cm.} .\\ \\ \tt\mapsto -\frac{1}{20 \: cm} - \frac{1}{u} = - \frac{1}{30 \: cm}. \\ \\ \tt\mapsto - \frac{1}{u} = - \frac{1}{30 \: cm} + \frac{1}{20 \: cm} . \\ \\ \tt\mapsto \frac{1}{u} = \frac{1}{30 \: cm} - \frac{1}{20 \: cm} . \\ \\ \tt\mapsto \frac{1}{u} = \frac{2 - 3}{60 \: cm} . \\ \\ \tt\mapsto \frac{1}{u} = \frac{-1}{60 \: cm} . \\ \\ \tt\mapsto \frac{\cancel1}{u} = -\frac{\cancel1}{60 \: cm} \\ \\ \tt\mapsto u = -60 \: cm.$

$\tt\therefore$ The object should be placed 60 cm in front of the lens.