Question

what will be the range of value of M for erect and diminished image formed by a convex mirror of a real object
a. m>1
b. 0 c. -1 d. m<-1

Answer 1

Explanation:

option a is the correct one .

Answer 2

The correct option is (a) i.e. m>1.

Explanation:

The mirror formula is given by :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

u is the object distance

v is the image distance

f is the focal length

Magnification of convex mirror is given by :

$m=\dfrac{-v}{u}=\dfrac{h'}{h}$

h'is the height of image

h is the height of object

On solving above two formulas we get :

$m=\dfrac{f}{f+u}$

In the formula for magnification, the numerator is lesser than the denominator for all values of |u|>0 and both the numerator and denominator are always positive numbers. As a result, the magnification is a positive number and is always lesser than for erect and diminished image formed by a convex mirror of a real object. So, the correct option is (a) i.e. m>1.

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Magnification

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