Answer to Question #260257 in Optics for Nbj

Question

Q3. An object is placed 25 cm in front of the convex lens whose focal length is 10 cm.

(a) Find the image distance and where it is located

Now a concave lens is placed 6 cm behind the convex lens and a new image is formed 20

cm behind the convex lens. Draw a rough diagram and indicate the length measurements

(b) Calculate the focal length of the concave lens (hint: the image formed by the

convex lens will be the object to the concave lens)

Answer 1

$3. (a) \\ f\,=\,10cm \\ u\,=\,25cm\\ \frac{1}{f}\,=\,\frac{1}{v}\,+\,\frac{1}{u}\\ \frac{1}{v}\,=\,\frac{1}{f}\,-\,\frac{1}{u}\\ \frac{1}{v}\,=\,\frac{1}{10}\,-\,\frac{1}{25}\\ \frac{1}{v}\,=\,\frac{5-2}{50}\\ \frac{1}{v}\,=\,\frac{3}{50}\\ v\,=\,{16.67cm}\\ image\,distance\,=\,{16.67cm}\\ {3.b}\\ {for}\,convex\,lense\\ \frac{1}{f}\,=\,\frac{1}{v}\,+\,\frac{1}{u}\\ \frac{1}{10}\,=\,\frac{1}{v}\,+\,\frac{1}{20}\\ \frac{1}{v}\,=\,\frac{1}{10}\,-\,\frac{1}{20}\\ \frac{1}{v}\,=\,\frac{2-1}{20}\\ \frac{1}{v}\,=\,\frac{1}{20}\\ {v}\,=\,{20cm}\\ so\,u\,for\,concave\,lens\,=\,{20-6}\,cm\\ {u}\,=\,{14cm}\\ \frac{1}{f}\,=\,\frac{1}{v}\,-\,\frac{1}{u}\\ \frac{1}{f}\,=\,\frac{1}{\alpha}\,-\,\frac{1}{14}\\ \frac{1}{f}\,=\,\frac{1}{-14}\\ {f}\,=\,{-14cm}\\$