The distance between an object and a screen is 100 cm. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is 40 cm. If the power of the lens is close to ( N/100 )D where N is an integer, the value of N is _________.

Solution

Physics sir · 4 steps

The two lens positions just swap object and image distances, so the 100 cm and 40 cm fix both of them.

1. Swap idea. Light paths are reversible. If the lens works with object distance x and image distance y, it also works with y and x. The two positions are x from the object and y from the object, so they are y - x apart.
2. Find the distances. x + y = 100 cm and y - x = 40 cm. Adding gives y = 70 cm, and then x = 30 cm.

   1/f = 1/v - 1/u = 1/70 + 1/30 = 10/210 ⇒ f = 21 cm

3. Cross-check. The displacement-method formula f = (D² - d²)/4D = (10000 - 1600)/400 = 8400/400 also gives f = 21 cm.
4. So. P = 100/(f (in cm)) = 100/21 ≈ 4.76 D = 476/100 D. So N = 476.

The common slip: 119. Dropping the 4 in f = (D² - d²)/4D gives f = 84 cm, so P ≈ 1.19 D and N = 119, not 476.

The idea to keep. In the displacement method, the two sharp-image positions are conjugate: u and v swap. Use f = (D² - d²)/4D, and put f in metres before you take the power.

Answer from the NTA final answer key. Solution by PrepWiser.