CBSE Class 10 Answered
12 cm is the radius of curvature of concave mirror. The image obtained by it is real and has magnification of the 3/2. Find object distance
Asked by navdurgacrane1997 | 16 Jul, 2018, 20:50: PM
focal length f of concave mirror = 1/2 of radius of curvature = 12/2 = 6 cm .
magnification of mirror, m = -v/u = -1.5 ..............(1)
where v is image distance and u is object distance
(it is given that image is real. In concave mirror real images are inverted, hence -ve sign i.e. -1.5 is used in eqn.(1) )
from eqn.(1) , v = 1.5 u ................(2)
we have mirror eqn., ![begin mathsize 12px style 1 over v space plus space 1 over u space equals space 1 over f space................. left parenthesis 3 right parenthesis end style](https://images.topperlearning.com/topper/tinymce/cache/ec3a94a6ebc3e395103049e09a504c6d.png)
![begin mathsize 12px style 1 over v space plus space 1 over u space equals space 1 over f space................. left parenthesis 3 right parenthesis end style](https://images.topperlearning.com/topper/tinymce/cache/ec3a94a6ebc3e395103049e09a504c6d.png)
Substituting value of f and using eqn.(2), eqn.(3) becomes : ![begin mathsize 12px style fraction numerator 2 over denominator 3 u end fraction plus 1 over u space equals space 1 over 6 space............... left parenthesis 4 right parenthesis end style](https://images.topperlearning.com/topper/tinymce/cache/3bb9d924927370f3a1604293fb2b14fc.png)
![begin mathsize 12px style fraction numerator 2 over denominator 3 u end fraction plus 1 over u space equals space 1 over 6 space............... left parenthesis 4 right parenthesis end style](https://images.topperlearning.com/topper/tinymce/cache/3bb9d924927370f3a1604293fb2b14fc.png)
solving for u in the eqn.(4), we get u = 10 cm
Answered by Thiyagarajan K | 17 Jul, 2018, 16:46: PM
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