Question 6(c)A lens of focal length 20 cm forms an inverted image at a distance 60 cm from the lens. [4](i) Identify the lens.(ii) How far is the lens present in front of the object?(iii) Calculate the magnification of the image.

Question

Accepted Answer

(i) The lens used is convex lens.(ii) Given, f = +20 cm, v = +60 cm, u = ?From lens formula,1f\dfrac{1}{f}f1​ = 1v\dfrac{1}{v}v1​ - 1u\dfrac{1}{u}u1​Substituting the values in the formula,120=160−1u⇒1u=160−120⇒1u=1−360⇒1u=−260⇒u=−30cm\dfrac{1}{20} = \dfrac{1}{60} - \dfrac{1}{u} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{1}{60} - \dfrac{1}{20} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{1 - 3}{60} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{- 2}{60} \[0.5em] \Rightarrow u = - 30 cm \[0.5em]201​=601​−u1​⇒u1​=601​−201​⇒u1​=601−3​⇒u1​=60−2​⇒u=−30cmHence, the lens is 30 cm in front of the object(iii) Magnification m = vu\dfrac{v}{u}uv​Substituting the values in the formula above we get,m=60−30=−2m = \dfrac{60}{- 30} = -2m=−3060​=−2Thus, magnification = 2 times

Solved 2020 Question Paper ICSE Class 10 Physics — Question 6

Question 6(c)