Dear student formula of shift is x=d2-4fd ( d is the distance of the object and the screen )30=d2-4fd 1 formula of magnification in case of lens displacement method is m=d+d2-4fdd-d2-4fd3=d+d2-4fdd-d2-4fd3d-3d2-4fd=d+d2-4fd2d=4d2-4fdd=2d2-4fdd2=4(d2-4fd)d2=4d2-16fd3d=16fd=16f330=d2-4fdsquaring both sides900=d2-4fd900=1629f2-4f×16f3900=1629f2-4f×16f3=f2(1629-643)=649f2f2=900×964f=900×964=3×308=11 25cmRegards

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NO LINKS PLEASE By placing a convex lens between an object and a screen, an image magnified 3 times of its object is obtained on the screen. By shift. ing the lens through a distance of 30 cm towards the screen, again an image is obtained on the screen. What is the focal length of the lens?

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f o r m u l a o f s h i f t i s x = \sqrt{d^{2} - 4 f d} (d i s t h e d i s \tan c e o f t h e o b j e c t a n d t h e s c r e e n) 30 = \sqrt{d^{2} - 4 f d} . . . 1 f o r m u l a o f m a g n i f i c a t i o n i n c a s e o f l e n s d i s p l a c e m e n t m e t h o d i s m = \frac{d + \sqrt{d^{2} - 4 f d}}{d - \sqrt{d^{2} - 4 f d}} 3 = \frac{d + \sqrt{d^{2} - 4 f d}}{d - \sqrt{d^{2} - 4 f d}} 3 d - 3 \sqrt{d^{2} - 4 f d} = d + \sqrt{d^{2} - 4 f d} 2 d = 4 \sqrt{d^{2} - 4 f d} d = 2 \sqrt{d^{2} - 4 f d} d^{2} = 4 (d^{2} - 4 f d) d^{2} = 4 d^{2} - 16 f d 3 d = 16 f d = \frac{16 f}{3} 30 = \sqrt{d^{2} - 4 f d} s q u a r i n g b o t h s i d e s 900 = d^{2} - 4 f d 900 = \frac{16^{2}}{9} f^{2} - 4 f \times \frac{16 f}{3} 900 = \frac{16^{2}}{9} f^{2} - 4 f \times \frac{16 f}{3} = f^{2} (\frac{16^{2}}{9} - \frac{64}{3}) = \frac{64}{9} f^{2} f^{2} = \frac{900 \times 9}{64} f = \sqrt{\frac{900 \times 9}{64}} = \frac{3 \times 30}{8} = 11.25 c m R e g a r d s

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