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CBSE Class 10 Answered

An object is placed at distance 15 cm from a concave mirror of focal length 10 cm. Find position, nature and magnification of image.
Asked by rishikadubey462 | 13 Sep, 2021, 08:09: AM
answered-by-expert Expert Answer
object distance, u = - 15 cm 
image distance, v = ?
focal length = -10 cm (focal length of concave mirror)
 
1 over f equals 1 over u plus 1 over v fraction numerator 1 over denominator negative 10 end fraction equals fraction numerator 1 over denominator negative 15 end fraction plus 1 over v 1 over v space equals fraction numerator 1 over denominator negative 10 end fraction space minus space space open parentheses fraction numerator 1 over denominator negative 15 end fraction close parentheses 1 over v space equals space fraction numerator 1 over denominator negative 10 end fraction plus fraction numerator 1 over denominator negative 15 end fraction 1 over v space equals space fraction numerator negative 15 space plus space left parenthesis negative 10 right parenthesis over denominator 150 space end fraction 1 over v equals space fraction numerator negative 5 over denominator 150 space end fraction v space equals space minus space 30 space c m space  
Negative sign indicate that the image will be formed on same side as that of object. 
Thus, image is real and inverted. 
 
m = - v/u = - (-30/-15) = - 2 
 
Thus magnification of the image is -2. 
Negative sign indicates that the image formed is inverted and value 2 inidicates that the image formed is twice the size of the object. 
Thus,nature of image is real, inverted and magnified. 
Answered by Shiwani Sawant | 20 Sep, 2021, 01:39: PM

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