Question

The focal length of a concave mirror is f and the

distance of the object from the principal focus is a. The magnitude of magnification obtained will be-​

Answer 1

Answer:

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Explanation:

As per the sign convention for mirror, the focal length is taken as negative as it is measured opposite to the direction of light. Hence, focal length f = -f. The distance of the object from the principal focus is p. Hence, the distance of object from the mirror [ u ] = p+ f.

OR

Distance of object is u=−(f+p)

Distance of object is u=−(f+p)v1+u1=f1 gives:

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1or, v1=−(f+p)×fp

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1or, v1=−(f+p)×fpor, v=−p(f+p)×f      (-ve sign indicates image is real) 

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1or, v1=−(f+p)×fpor, v=−p(f+p)×f      (-ve sign indicates image is real)    Magnification =−uv 

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1or, v1=−(f+p)×fpor, v=−p(f+p)×f      (-ve sign indicates image is real)    Magnification =−uv            =−p×(f+p)(f+p)×f   

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1or, v1=−(f+p)×fpor, v=−p(f+p)×f      (-ve sign indicates image is real)    Magnification =−uv            =−p×(f+p)(f+p)×f              =−pf  (-ve sign indicates inverted)

Distance of object is u=−(f+p)v1+u1=f1 gives:v1−f+p1=−f1or, v1=−f1+f+p1or, v1=−(f+p)×fpor, v=−p(f+p)×f      (-ve sign indicates image is real)    Magnification =−uv            =−p×(f+p)(f+p)×f              =−pf  (-ve sign indicates inverted)    So, ratio of size of image to that of object is: −pf

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