The distance between on object and a scren is 100 cm. A lens can produce real image of the object on

Q: Why are there exactly two positions for the lens to form an image?

This is due to the principle of reversibility of light. The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ is symmetric with respect to object and image distances. If a lens forms an image at distance $v$ for an object at $u$, it will also form an image at distance $u$ if the object is placed at $v$. These are called conjugate positions.

Q: What happens if the distance $D$ is exactly equal to $4f$?

If $D = 4f$, the quadratic equation for the lens position has equal roots. This means the two positions merge into a single position exactly midway between the object and the screen. In this case, $d = 0$.

Q: Why did we have to convert the focal length to meters?

The unit of optical power is the Diopter (D), which is defined as the reciprocal of the focal length measured strictly in meters ($1\text{ D} = 1\text{ m}^{-1}$). Using centimeters will give you an incorrect power value.

Q: Can we use the displacement method for a concave lens?

No, the displacement method requires the formation of a real image on a screen. A concave lens always forms virtual, erect, and diminished images for real objects, so it cannot project an image onto a screen.

Q: How is the formula $f = \frac{D^2 - d^2}{4D}$ actually derived?

It comes from setting up the lens formula. Let the object distance be $x$, then the image distance is $D - x$. Applying the sign convention, $u = -x$ and $v = D - x$. The lens formula becomes $\frac{1}{D-x} + \frac{1}{x} = \frac{1}{f}$. This forms a quadratic equation $x^2 - Dx + Df = 0$. The difference between the two roots of this equation is the displacement $d$, which leads to the final formula.

Question

The distance between on object and a scren is $100	ext{ cm}$. 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 $40	ext{ cm}$. If the power of the lens is close to $\left(\frac{N}{100}ight)	ext{ D}$, where $N$ is an integer. Then, the value of $N$ is ……… .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an optical bench setup where an object and a screen are fixed at a certain distance apart. This distance is given as $D = 100	ext{ cm}$. When we move a convex lens between the object and the screen, we discover a fascinating phenomenon: there are exactly two distinct positions of the lens where a sharp, real image is formed on the screen.

The distance between these two specific lens positions is denoted as $d$, and in our case, $d = 40	ext{ cm}$. This entire scenario is a classic application of the **Displacement Method**, a highly reliable experimental technique used to determine the focal length of a convex lens.

### The Master Equation

For the displacement method, the relationship between the focal length $f$, the distance between the object and screen $D$, and the displacement of the lens $d$ is given by a direct and elegant formula:

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

Let's carefully substitute the values we have into this master equation. We plug in $D = 100$ and $d = 40$:

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

### Executing the Calculation

Now, we perform the arithmetic. It is crucial to avoid silly mistakes here. Squaring the terms gives us:

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

Subtracting the numerator values:

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

Dividing $8400$ by $400$, we find the focal length of the convex lens:

$$f = 21	ext{ cm}$$

### The Power Trap

The question doesn't stop at the focal length; it asks for the **optical power** of the lens. The formula for power is $P = \frac{1}{f}$. However, there is a massive trap here: to calculate power in Diopters (D), the focal length **must** be in meters.

Let's convert our focal length:

$$f = 21	ext{ cm} = 0.21	ext{ m}$$

Now, we substitute this into the power formula:

$$P = \frac{1}{0.21}$$

To simplify, we remove the decimal by multiplying the numerator and denominator by 100:

$$P = \frac{100}{21}	ext{ D}$$

If we evaluate this fraction, it comes out to approximately $4.76	ext{ D}$.

### Final Comparison

The problem states that the power of the lens is close to $\left(\frac{N}{100}ight)	ext{ D}$. We can rewrite our calculated decimal power in a similar fractional format:

$$4.76	ext{ D} = \frac{476}{100}	ext{ D}$$

By comparing our result with the given expression $\frac{N}{100}$, it becomes immediately clear that:

**$$N = 476$$**

This is our final integer answer. Always remember the critical condition for the displacement method: a real image on the screen is only possible if the distance $D$ is greater than or equal to $4f$. If $D < 4f$, no such lens positions will exist!

Enter Numerical Value:

Visualized Solution

Visualizing the Displacement Method

The Master Formula

Substituting the Values

Calculating Focal Length

Optical Power Formula

Calculating Power

Finding N

The Way Forward

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Executing the Calculation

The Power Trap

Final Comparison

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