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
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](https://images.topperlearning.com/topper/tinymce/cache/9fa975a3c2994655b76d8f703239ec55.png)
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, 13:39: PM
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