Question

A 10 Milli metre long awl pin is placed vertically in front of a concave mirror. A 5 Milli metre long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is..... with clear explanation​

Answer 1

Given:

hi= 5mm = -0.5cm

ho= 10mm = 1cm

V = -30cm

Sol.:

Magnification=hi/ho

= -0.5/1 = -1/2

Also, magnification = -v/u

=-(-30)/u

==>-1/2=-(-30)/u

therefore , u=-60cm

Now by using Mirror formula,

1/f =1/v+1/u

1/f=(-1/30)+(-1/60)

1/f=(-1/20)

Therefore f = -20 cm

Hope it helps u

Answer 2

We are given ,

Height of the object as 10mm

Height of the image formed as 5mm

Image distance i.e v = -30cm= -300mm

To find : u or object distance and f or focal length

Solution : we know that m= -v÷u = hi÷ho

So, -v÷u=hi÷ho

So, -(-300)÷u= 5÷10

So,u = +300×2= +600mm

But. Acc. To sign convention u can't be +ve

So, u = -600mm = -60cm

Now

Mirror formula =1÷f = 1÷v +1÷u

So , 1÷f = 1÷[-30] + 1÷ (-60)

Taking LCM

1÷f = -2-1÷60

= -3÷60

Therefore , f = -60÷3

So.f= -20 cm