Question

A Concave mirror gives virtual,refract and enlarged image of the object but image of smaller size then object is:
a) At infinity
b) Between F and C
c) Between P and F
d) At E

KINDLY EXPLAIN WHY

AND ALSO WHAT IS THIS E POSITION IN MIRROR?​

Answer 1

Then the object is between p and f

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if it is convex

Answer 2

Answer:

C) Between P and F is the correct answer.

Explanation:

We know that the magnification produced by a mirror is given by,

m=−v/u  (1)

Now, from the mirror formula we have,

1/v + 1/u = 1/f

Multiplying with −u on both the sides, we get

−u/v + −u/u = −u/f

⇒−u/v = 1−u/f

Taking LCM on the RHS, we have

−u/v = f−u/f

Taking the reciprocals of both the sides, we get

−v/u = f/f−u

From (1)

m = f/f−u    (2)

Now, we know that according to the Cartesian sign convention, both the object distance and the focal length of a concave mirror are taken as negative. This means that we can write

f = −|f|    (3)

Also

u = −|u|    (4)

Putting (3) and (4) in (2) we get

m = −|f|−|f|+|u|

Multiplying the numerator and the denominator by −1 we get

m = |f| |f| − |u|

According to the question, the image formed is virtual, erect and enlarged. Since the virtual and erect, the magnification is positive. This means that

|f| |f| − |u| > 0

⇒|f| > |u|

Or

|u|<|f|

So the object distance is less than the focal length of the mirror.

This means that the object must be kept between the pole and the focus of the concave mirror.

Hence, the correct answer is option (C).