JEE PYQ: Ray Optics Question 19
Question 19 - 2020 (04 Sep Shift 2)
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 $\left(\frac{N}{100}\right)D$ where $N$ is an integer, the value of $N$ is _________.
Show Answer
Answer: (476.19)
Solution
Given, Distance between an object and screen, $D = 100$ cm Distance between the two position of lens, $d = 40$ cm Focal length of lens, $f = \frac{D^2 - d^2}{4D} = \frac{100^2 - 40^2}{4(100)} = \frac{(100 + 40)(100 - 40)}{4(100)} = 21$ cm
Power, $P = \frac{1}{f} = \frac{100}{21} = \frac{N}{100}$ $\therefore N = 476.19$
BETA
