Question

6. A concave mirror of focal length 20 cm produces an image twice the height of the object
image is real, then the distance of the object from the mirror is
(d) 30 cm​

Answer 1

Given :-

• focal length of concave mirror , f = -20 cm
• height of image (h) is twice the height of object (h')
• image formed is real

To find :-

• distance of object from the mirror = ?

Formula required :-

• formula for magnification produced by mirror

$\pink{\bigstar}\boxed{\sf{m=\dfrac{-v}{u}=\dfrac{h}{h\;'}}}$

(where m is magnification of produced by mirror , u is the position of object , v is the position of image , h is height of image , h' is height of object)

• Mirror formula

$\pink{\bigstar}\boxed{\sf{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}$

(where f is the focal length of mirror , u is position of object , and v is position of image)

Solution :-

Since, Image formed is real

therefore , image formed will also be inverted

and given that height of image is twice the height of object

so,

$\implies\sf{height\:of\:image=-2(height\:of\:object)}\\\\\implies\red{\sf{h=-2\;h'\;\;\;.......eqn(1)}}$

Now, using formula for magnification

$\implies\sf{\dfrac{-v}{u}=\dfrac{h}{h'}}\\\\\sf{using\:eqn\:(1)}\\\\\implies\sf{\dfrac{-v}{u}=\dfrac{-2h'}{h'}}\\\\\implies\sf{\dfrac{-v}{u}=\dfrac{-2}{1}}\\\\\implies\red{\sf{v=2u\;\;\;\;....eqn(2)}}$

Using mirror formula

$\implies\sf{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}\\\\\implies\sf{\dfrac{1}{-20}=\dfrac{1}{v}+\dfrac{1}{u}}\\\\\sf{using\:eqn\:(2)}\\\\\implies\sf{\dfrac{1}{-20}=\dfrac{1}{2u}+\dfrac{1}{u}}\\\\\implies\sf{\dfrac{1}{-20}=\dfrac{1+2}{2u}}\\\\\implies\sf{2u=3(-20)}\\\\\implies\underbrace{\boxed{\red{\sf{u=-30\;cm}}}}$

Hence,

Distance of object from the mirror is 30 cm .

(negative sign in u = -30 cm is because, according to the sign convention object is always kept in the negative principal axis)