Question 5(i)An object is placed in front of a concave lens at a distance of 45 cm from it. If its image is formed at a distance of 30 cm from the lens, calculate the focal length of the lens.

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(a) Given,Object distance (u) = -45 cmImage distance (v) = -30 cmFrom lens formula,1f=1v−1u1f=1−30−1(−45)1f=−130+1451f=−3+2901f=−190f=−901f=−90 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \[1em] \dfrac{1}{\text f} = \dfrac{1}{-30}-\dfrac{1}{(-45)}\[1em] \dfrac{1}{\text f}=-\dfrac{1}{30}+\dfrac{1}{45}\[1em] \dfrac{1}{\text f} = \dfrac{-3+2}{90}\[1em] \dfrac{1}{\text f} = -\dfrac{1}{90} \[1 em] \text f = -\dfrac{90}{1}\[1 em] \text f= - 90 \text { cm}f1​=v1​−u1​f1​=−301​−(−45)1​f1​=−301​+451​f1​=90−3+2​f1​=−901​f=−190​f=−90 cm

Solved 2025 Question Paper ICSE Class 10 Physics — Question 4

Question 5(i)