A rod of length 10 cm lies along the principal axis of a concave mirror of focal . . .

Question : A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is [AIPMT 2012]
(1) 10 cm
(2) 15 cm
(3) 2.5 cm
(4) 5 cm

Doubt by Girisha

Solution : 


Given:
Length of rod = 10 cm
Focal length of concave mirror (f) = 10 cm
Distance of the nearer end from the mirror (Pole) = u1 = 20 cm
Distance of the farther end from the mirror (Pole) = u2 = 10+20 = 30 cm 

We know, 
Mirror Formula is given by 
1/v + 1/u = 1/f

Now, position of the image of the neare end from the pole or mirror (v1)
u1 = −20 cm, f1 = −10 cm
1/v1 + 1/u1 = 1/f1
1/v1 = 1/f1 - 1/u1
1/v1 = −1/10 + 1/20
1/v1 = −1/20
v1 = −20 cm

Now, position of the image of the farther end from the pole or mirror (v2)
u2 = −30 cm, f1 = −10 cm
1/v2 + 1/u2 = 1/f1
1/v2 = 1/f1 - 1/u2
1/v2 = −1/10 + 1/30
1/v2 = −2/30
v2 = −15 cm

Length of image
= Difference between the image positions
= |-20| − |-15|
= 5 cm

Hence, (4) 5 cm, is the correct option. 

Alternate Method 

Given:
Length of rod = 10 cm
Focal length of concave mirror (f) = 10 cm
Distance of the nearer end from the mirror (Pole) = u1 = 20 cm
Distance of the farther end from the mirror (Pole) = u2 = 10+20 = 30 cm 

We know, 
Mirror Formula is given by 
1/v + 1/u = 1/f
1/v = 1/f - 1/u
1/v = (u-f) / fu
v = fu / (u-f)
Here u and f both are -ve 
so putting f = -f and u = -u
v = fu / (-u+f)
v = fu / (f-u)

Now, position of the image of the neare end from the pole or mirror (v1)
v1 = fu1 / (f-u1)
v1 = (10)(20) / (10-20)
v1 = 200/(-10)
v1 = -20 cm 

Now, position of the image of the farther end from the pole or mirror (v2)

v2 = fu2 / (f-u2)
v2 = (10)(30) / (10-30)
v2 = 300/(-20)
v2 = -15 cm 


Length of image
= Difference between the image positions
= |-20| − |-15|
= 5 cm

Hence, (4) 5 cm, is the correct option.