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CBSE Class 10 Answered

a thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. calculate the magnification.
Asked by apurva151 | 14 Jun, 2015, 06:25: PM
answered-by-expert Expert Answer
 
Given that Rod, AB is placed along the optical axis.
The image of the rod will touch the rod, AB only when the end Cof the rod is placed at the centre of curvature of the concave mirror.
PC = 2f and CA = f/ 3
The image of the end C of the rod will be formed at C itself in this situation
At A side of the rod:
u = PA = PC - AC  = 2f - (f/3)  = 5f/3
Using the mirror formula, we get:
begin mathsize 12px style 1 over straight f equals 1 over straight v plus space 1 over straight u Or comma space 1 over straight v equals 1 over straight f minus space 1 over straight u space space space space space space space space space space equals space 1 over straight f minus space fraction numerator 1 over denominator left parenthesis begin display style bevelled fraction numerator 5 straight f over denominator 3 end fraction end style right parenthesis end fraction space space space space space space space space space space equals space 1 over straight f minus space fraction numerator 3 over denominator 5 straight f end fraction space equals space fraction numerator 2 over denominator 5 straight f end fraction Or comma space straight v space equals fraction numerator 5 straight f over denominator 2 end fraction Thus comma space the space image space of space straight A space of space the space rod space is space formed space at space space straight A apostrophe space which space is space at space straight a space distance 5 straight f divided by 2 space from space the space pole space of space the space mirror. So comma space PA apostrophe space equals 5 straight f divided by 2 Length space of space the space image space of space the space rod comma space CA apostrophe space equals space PA apostrophe space minus space PC space equals space left parenthesis 5 straight f divided by 2 right parenthesis space minus space 2 straight f space equals space straight f divided by 2 Magnification comma space straight M space equals space fraction numerator Size space of space image space over denominator Size space of space object end fraction Or comma space straight M space equals space fraction numerator CA apostrophe over denominator CA end fraction straight M equals space fraction numerator straight f divided by 2 over denominator straight f divided by 3 end fraction space equals space 3 over 2 equals space 1.5 end style
Answered by Yashvanti Jain | 15 Dec, 2017, 03:35: PM

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