Question

 An object is placed in front of a concave mirror of focal length 12cm at a 
      distance 30cm from the mirror  , Find 
   a) position of the image 
   b) Nature of the ima

Answer 1

Answer:

1/u + 1/v = 1/f

1/v = 1/-12+1/30 (use sign conventions)

1/v= -1/20

v= -20

Answer 2

Given :

In concave mirror,

Focal length = -12cm

Object distance = -30cm

Note : In concave mirror focal length is always negative.

To find :

(a) position of the image 

(b) Nature of the image

Solution :

(a) Position of image

Using lens formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

$\boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

where,

• v denotes Image distance
• u denotes object distance
• f denotes focal length

now, substituting all the given values,

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{( - 30)} = \dfrac{1}{( - 12)} }$

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{30} = - \dfrac{1}{12} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{12} + \dfrac{1}{30} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{ - 5 + 2}{60} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{ - 3}{60} }$

$\leadsto{ \sf v = - \dfrac{60}{3} }$

$\leadsto{ \sf v = - 20 \: cm }$

thus, the position of image is - 20cm.

(b) Nature of image

The minus signhere shows that the image is formed below the principle axis. This is the image is real and inverted.