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CBSE Class 12-science Answered

Use the mirror formula to show that  a)an object placed between f and 2f of a concave mirror produces a real image  beyond 2f b)a convex mirror always produces a virtual image independent of the location of the object c)an object placed between the pole and focus of a concave mirror produces virtual and enlarged image
Asked by vasturushi | 02 Feb, 2018, 06:58: PM

(i)

For a concave mirror, focal length (f) is negative, f<0

When an object s placed on the left side of the mirror, the object distance (u) is negative, u<0

Using lens formula

1/ v  - 1/ u = 1/f

1/v = 1/f -1/u  ......(i)

The object lies between f and 2f

2f<u<f

1/2f>1/u>1/f

1/f -1/2f<1/f-1/u<0 .......(ii)

Using equation (i) we get

1/2f< 1/v<0                if, 1/v is negative

or, 1/2f<1/v

or, 2f>v

-v>-2f

Therefore, the image lies beyond 2f

(ii)

For a convex mirror, the focal length (f)>0 and u<0

1/v +1/u = 1/f

1/v = 1/f -1/u

Using equation (ii) we get

1/v<0

v>0

Hence a convex mirror is always produces a virtual image.

(iii)

For a convex mirror, the focal length (f)>0 or positive and u<0

Using les formula

1/v +1/u = 1/f

1/v = 1/f -1/u

Since u<0

Then , 1/v>1/f

v>f

(iv)

For a concave mirror, f is negative, i.e, f<0 and u<0

It is placed between f and pole

Therefore,

f>u>0

1/f<1/u<0

1/f-1/u<0

For negative distance v, using lens formula

1/v +1/u =1/f

1/v = 1/f -1/u

1/v<0

v>0

The image is formed on the  right side of the mirror.

Hence, it is virtual image

For u<0 and v<0

Then,

1/u>1/v

v>u

Answered by Abhijeet Mishra | 04 Feb, 2018, 04:19: PM

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