Question

A concave mirror produces three times magnified real image of an object placed at 10cm in front of it . Where is the image located ??​

Answer 1

★Figure refer to attachment

Given

A concave mirror produces three times magnified real image of an object placed at 10cm in front of it

To find

Where is the image located

Solution

• Magnification (m) = - 3
• Object distance (u) = - 10cm
• Image distance (v) = ?

$\implies\sf m=\dfrac{-v}{u} \\ \\ \\ \implies\sf -3 =\dfrac{-v}{-10} \\ \\ \\ \implies\sf v = -30cm$

Now, apply Mirror Formula

$\implies\sf \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{-1}{30}+\dfrac{-1}{10}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{-1-3}{30}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{-4}{30}=\dfrac{1}{f} \\ \\ \\ \implies\sf f=\dfrac{-30}{4}=-7.5cm$

Hence,

★ Image distance = -30cm

★ Focal length = - 7.5cm

Nature of image

• Nature = Real and virtual
• Size = Enlarged
• Position = Beyond centre of curvature

Answer 2

GiVeN :-

A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it.

• Object distance, u = -10 cm

• Magnification = negative

• (for producing a real image)

• Image size($h_i$) = 3 × Object size($h_o$)

To DeTeRmInE :-

The image distance(v).

AcKnOwLeDgEmEnT :-

○ Magnification :-

$\sf{\dfrac{-v}{u}=\dfrac{h_i}{h_o} }$

SoLuTiOn :-

Putting the values :-

$\sf{\dfrac{-v}{u}=\dfrac{h_i}{h_o} }$

$\sf{\longmapsto -(\dfrac{-v}{-10})=\dfrac{3h_o}{h_o} }\\\\\\\sf{\longmapsto \dfrac{v}{-10}=3 }\\\\\\\boxed{\sf{\longmapsto v=-30\ cm }}$

Therefore, the image is located 30 cm in front of the mirror.

(Negative sign is used when an object or an image is on the front of the mirror, i.e. left side of the mirror.)

$\rule{150}2$