a concave lens forms An erect image of 1/3 size of the object which is

1.	a concave lens forms An erect image of 1/3 size of the object which is placed at a distance 30 cm in front of a lens,find position of image, focal length of the lens
Answer» ANSWER: The FOCAL LENGTH is 15 cm and the image is formed at 10 cm from the lens. Explanation: Given that, Distance of the OBJECT u = -30 cm Magnification $m =\dfrac{h'}{h}= \dfrac{1}{3}$ We know that, The magnification is $m = \dfrac{h'}{h}=\dfrac{v}{u}$ $\dfrac{1}{3}=\dfrac{v}{-30}$ $v = -10 cm$ The image is formed at 10 cm from the lens. Using lens's formula $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ $\dfrac{1}{f}=\dfrac{1}{-15}$ $f = -15 cm$ Hence, The focal length is 15 cm and the image is formed at 10 cm from the lens.