Question

An object and it’s image by a mirror of focal length ‘f’ have distances from the focus in the ratio 1
: 4, the object distance is
a) f/m b) 2f c) 3f / 2 d) 2f / 3

Answer 1

Answer:

( c ) 3f / 2

Step-by-step explanation:

Let the object distance of the mirror be 'u' and image distance be 'v' and focal length be 'f'

Ratio of distances from the focus to object and image=  1 : 4

Let the distances from focus to object and image be 'x' and '4x'  respectively

And hence the object distance and image distance will be :

• u = f + x
• v = f + 4x

Using mirror formula

$\Rightarrow \sf \dfrac{1}{f} =\dfrac{1}{v} + \dfrac{1}{u}\\\\\\\Rightarrow \sf \dfrac{1}{f} =\dfrac{1}{f + x} + \dfrac{1}{f+4x}\\\\\\\Rightarrow \sf \dfrac{1}{f} -\dfrac{1}{f+x} + \dfrac{1}{f+4x}\\\\\\ \Rightarrow \sf \dfrac{f+x-f}{f(f+x)} = \dfrac{1}{f+4x}\\\\\\ \Rightarrow \sf \dfrac{x}{f^2+fx} = \dfrac{1}{f+4x}$

⇒ x( f + 4x ) = f² + fx

⇒ fx + 4x² = f² + fx

⇒ 4x² = f²

⇒ x² = f² / 4

Taking square root on both sides

⇒ x = √( f/2 )²

⇒ x = f/2

Object distance = v = f + x = f + f/2 = ( 2f + f ) / 2 = 3f / 2

Therefore the object distance is ( c ) 3f / 2.

Answer 2

Answer:

$\huge\mathcal{\pink{answer=}}\frac{3f}{2}$

Step-by-step explanation:

Let the object distance of the mirror 'u' , image distance 'v' and focal length 'f'.

Ratio of distance from the focus to the object and image =1:4

Let the distance from focus to object and image be 'x' and '4x' respectively

And hence the object distance and image distance will be,

● u=f+x

● v=f+4x

— using mirror formula;

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{f} = \frac{1}{f + x} + \frac{1}{f + 4x}$

$\frac{1}{f} - \frac{1}{f + x} + \frac{1}{f + 4x}$

$\frac{f + x - f}{f(f + x)} = \frac{1}{f + 4x}$

$x(f + 4x) = {f}^{2} + fx$

$fx + 4 {x}^{2} = {f}^{2} + fx$

$4 {x}^{2} = {f}^{2}$

${x}^{2} = \frac{ {f}^{2} }{4}$

Taking square roots both

$x = \sqrt{ { \frac{f}{2} }^{2} }$

$x = \frac{f}{2}$

Object distance =v=f+x=f+f/2=(2f+f)/2

=3f/2

So, the ANSWER is option (c) 3f/2