JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 1)
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 $$\left( {{N \over {100}}} \right)D$$ where N is an
integer, the value of N is _________.
Answer
476
Explanation
Using displacement method
$$f = {{{D^2} - {d^2}} \over {4D}}$$
Here, D = 100 cm
and d = 40 cm
$$f = {{{{100}^2} - {{40}^2}} \over {4(100)}} = 21\,cm$$
$$P = {1 \over f} = {{100} \over {21}}D$$
$${N \over {100}} = {{100} \over {21}}$$
$$N = 476$$
$$f = {{{D^2} - {d^2}} \over {4D}}$$
Here, D = 100 cm
and d = 40 cm
$$f = {{{{100}^2} - {{40}^2}} \over {4(100)}} = 21\,cm$$
$$P = {1 \over f} = {{100} \over {21}}D$$
$${N \over {100}} = {{100} \over {21}}$$
$$N = 476$$
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