Question

A distance of 100 cm exists between an object and a screen. For two different positions between the screen and the object, a lens can produce a true image of the object on the screen. There is a 40-centimeter space between these two positions. The value of N is ________ if the lens's power is close to (N/100)D, where N is an integer.

Answer 1

Step: 1 Given data:

Distance between an object and screen $\left(\mathrm{D}\right)=100$.

Distance between two points $\left(\mathrm{d}\right)=40$.

Power of lens $\approx \frac{\mathrm{N}}{100}$

Step: 2 Formula used:

$\mathrm{f}=\frac{{\mathrm{D}}^2-{\mathrm{d}}^2}{4\mathrm{D}}$.

$\mathrm{p}=\frac{1}{\mathrm{f}}$

Step: 3 Calculate N:

$$\begin{gathered} \mathrm{f}=\frac{{\mathrm{D}}^2-{\mathrm{d}}^2}{4\mathrm{D}}=\frac{{\left(100\right)}^2-{\left(40\right)}^2}{4\left(100\right)} \\ =\frac{10000-1600}{400}=\frac{100-16}{4} \\ =\frac{84}{4}=21 \end{gathered}$$

Since, $\mathrm{p}=\frac{1}{\mathrm{f}}$

$$\begin{gathered} \mathrm{p}=\frac{1}{21}=\frac{1}{21}\times \frac{100}{100}=\frac{4.76}{100} \\ \frac{\mathrm{N}}{100}=\frac{4.76}{100} \\ \mathrm{N}\approx 5 \end{gathered}$$