Question

A lens is constructed of a plastic whose index of refraction is 1.5. The lens is concave on the left side with a radius of 50cm and convex on the right side, with a radius of 25cm. A) Draw a diagram showing the lens. B) What will be the focal length of this lens? C) Is the lens converging or diverging? D) We now place an object 67cm to the left of the lens, make a new diagram showing the lens the object and the image. Calculate the position of the image produced by this object. Calculate the magnification of the image. Draw the principle rays on the diagram which will locate this image. Is the image virtual? E) Where should the object be placed to obtain a real image whose magnification is 2?

Answer 1

As, focal length = f

=> 1/f = (u - 1) * (1/R1 - 1/R2)

=> 1/f =   (1.5 - 1) * (1/50 + 1/25)

=> focal length = 33.33 cm

The lens is converging .

As, 1/u + 1/v = 1/f

=> 1/v = 1/33.33 - 1/67

=> v = 66.32 cm

Thus, the position of the image = 66.32 cm right of lens

Thus, magnification of the image =   - 66.32/67

                                                      = - 0.989

Image is real .

Here, 1/u + 1/2u = 1/33.33

=> u = 50 cm

Thus, the object be placed to obtain a real image whose magnification is 2 =   50 cm to the left of lens .