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. Find power of lens

1. What is happening?

A thin lens makes an image according to the lens formula

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

where

• $f$ = focal length of lens,
• $u$ = object distance from lens (taken negative in sign convention but we will work with magnitudes here),
• $v$ = image distance from lens.

In Bessel's method, the distance between the object and the screen is fixed and called $D$.

• First position of lens: object distance $u_1$, image distance $v_1$.
• Second position of lens: object distance $u_2$, image distance $v_2$.

Because the lens moves but object and screen stay, we always have

$u + v = D$

for both positions. A nice result from the algebra (derived below) is

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

where $d$ is the separation between the two sharp-image positions.

2. Why does that formula work?

1. Start with $u_1 + v_1 = D$ and $u_2 + v_2 = D$.
2. Because the lens shifts by $d$, one distance gets longer by $d$ while the other gets shorter by $d$.
3. Multiplying $u$ and $v$ for both cases and equating through the lens formula eliminates $u$ and $v$, finally giving the compact relation above.

3. What the student must do in exam

1. Identify $D$ (100 cm) and $d$ (40 cm).
2. Plug into $f = (D^2 - d^2)/(4D)$.
3. Convert $f$ from cm to metre.
4. Power $P$ is $1/f$ in dioptre (D).

Think of it like this

You have a torch (object) at one end of a 1-metre long bench and a white wall (screen) at the other end. You slide a magnifying glass (lens) somewhere in between so that a sharp bright spot of the torch appears on the wall. Surprisingly, you find two spots where the picture becomes sharp. Those two lens positions are 40 cm apart. From just these two distances, we can know how strong the magnifying glass is (its power), just like knowing the size of a cricket bat by two simple measurements.

Step-by-step calculation

Given: $D = 100\ \text{cm}$, $d = 40\ \text{cm}$

Formula from Bessel method

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

Substitute the numbers

$f = \frac{(100)^2 - (40)^2}{4 \times 100}$

$f = \frac{10000 - 1600}{400}$

$f = \frac{8400}{400}$

$f = 21\ \text{cm}$

Convert to metre

$f = 21\ \text{cm} = 0.21\ \text{m}$

Power of the lens

$P = \frac{1}{f\,(\text{in metre})}$

$P = \frac{1}{0.21} \approx 4.76\ \text{dioptre}$

[\boxed{P \approx 4.8\ \text{D}}]