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Wave Optics

Question
CBSEENPH12039270

Use the mirror equation 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 a virtual and enlarged image.

Solution

Mirror equation is given by,
space space space space space space 1 over straight f space equals space 1 over v plus 1 over u

rightwards double arrow space space space 1 over v space equals space 1 over f space minus space 1 over u 

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

For a real object i.e., which is on the left side of the mirror, 
For u between f and 2f implies that 1/u lies between 1/f and 1/2f
i.e.,   space space space space space fraction numerator 1 over denominator 2 straight f end fraction greater than fraction numerator 1 over denominator straight u space end fraction greater than 1 over straight f space left parenthesis as space straight u comma straight f space are space negative right parenthesis

rightwards double arrow space 1 over straight f minus fraction numerator 1 over denominator 2 straight f end fraction less than space 1 over straight f minus fraction numerator 1 over denominator straight u space end fraction less than space 0

rightwards double arrow space fraction numerator 1 over denominator 2 straight f end fraction less than space 1 over straight v less than 0

straight i. straight e. comma space space space space 1 over straight v space is space negative. space
Implies, v is negative and greater than 2f. Therefore, image lies beyond 2f and it is real.

b)
Focal length is positive for convex mirror, i.e., f > 0.

For a real object on the left side of the mirror, u is negative.
1 over straight f space equals 1 over v space plus space 1 over u
That is, 
1 over straight v space equals space 1 over straight f space minus space 1 over straight u

Since u is negative and f is positive so, 1/v should also be positive, so v must be positive.

Hence, image is virtual and lies behind the mirror.
c) 
Using the mirror equation, we have
1 over straight v space equals space 1 over straight f space minus space 1 over straight u

For space straight a space concave space mirror comma space straight f space is space negative semicolon space straight f space less than space 0

As space straight u space is space also space negative comma space so space straight f space less than space straight u less than 0

This space implies comma space

1 over straight f space minus space 1 over straight u greater than 0

Then space from space left parenthesis 1 right parenthesis comma space 1 over straight v space greater than thin space 0 space or space straight v space is space positive. space

straight i. straight e. comma space image space is space formed space on space the space right space
hand space side space and space is space virtual. space

Magnification comma space straight m space equals negative straight v over straight u space equals space minus fraction numerator straight f over denominator straight u minus straight f end fraction

As space straight u space is space negative space and space straight f space is space positive comma space

magnification comma space straight m space equals space fraction numerator vertical line straight f vertical line over denominator vertical line straight f vertical line space minus space vertical line straight u vertical line end fraction greater than thin space 1 
That is, the image is enlarged.

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