Question 17A concave lens has a focal length of 30 cm. Find the position and magnification (m) of the image for an object placed in front of it at a distance of 30 cm. State whether the image is real or virtual ?

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Accepted Answer

As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,f = - 30 cmu = - 30 cmSubstituting the values in the formula, we get,1v–1−30=1−301v+130=−1301v=−130−1301v=−1−1301v=−2301v=−115⇒v=−15\dfrac{1}{v} – \dfrac{1}{-30} = \dfrac{1}{-30} \[0.5em] \dfrac{1}{v} + \dfrac{1}{30} = - \dfrac{1}{30} \[0.5em] \dfrac{1}{v} = -\dfrac{1}{30} - \dfrac{1}{30} \[0.5em] \dfrac{1}{v} = \dfrac{-1-1}{30} \[0.5em] \dfrac{1}{v} = -\dfrac{2}{30} \[0.5em] \dfrac{1}{v} = -\dfrac{1}{15} \[0.5em] \Rightarrow v = -15v1​–−301​=−301​v1​+301​=−301​v1​=−301​−301​v1​=30−1−1​v1​=−302​v1​=−151​⇒v=−15Therefore, the image in formed 15 cm in front of the lens.ii) As we know,the formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​ Given,u = - 30 cmv = - 15 cmSubstituting the values in the formula, we get,m=−15−30m=0.5m = \dfrac{-15}{-30} \[0.5em] m = 0.5 \[0.5em]m=−30−15​m=0.5Therefore, the magnification is +0.5The image is virtual as the magnification is positive.

Refraction Through a Lens — Question 17

Question 17