In an optics experiment, with the position of the object fixed, a student varies the position of a c

Q: Why do we use $+v$ and $-u$ in the lens formula?

Because the object is real (placed in front of the lens, so $u$ is negative) and the image is real (formed on a screen behind the lens, so $v$ is positive). The graph plots the magnitudes of these distances.

Q: What does the curve in the graph represent?

The curve represents the relationship between the magnitude of the image distance $v$ and the magnitude of the object distance $u$ for a convex lens, given by $v = \frac{uf}{u-f}$.

Q: Why is the angle exactly $45^\circ$?

The $45^\circ$ angle is a specific condition given in the problem to find the point where the object and image distances are equal in magnitude ($v=u$).

Q: Can this point $P$ be achieved with a concave lens?

No, a concave lens always forms a virtual, erect, and diminished image for a real object, so the magnitude of the image distance $v$ is always less than the object distance $u$. They can never be equal except at the optical center.

Q: What is the magnification at point $P$?

Since the image is real and inverted, and $v=u$, the magnification is $m = -\frac{v}{u} = -1$. The image is the same size as the object.

Question

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $u$ and the image distance $v$, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of $45^\circ$ with the x-axis meets the experimental curve at $P$. The coordinates of $P$ will be

Options:

(A) $(2f, 2f)$

(B) $\left(\frac{f}{2}, \frac{f}{2}\right)$

(C) $(f, f)$

(D) $(4f, 4f)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in an optics lab, performing an experiment with a convex lens. You keep the object fixed and move the lens to different positions. For each position, you adjust a screen to catch a sharp, clear image.

You then plot a graph of the image distance $v$ against the object distance $u$. Both axes use the same scale. The resulting curve represents the relationship between $u$ and $v$ for this specific lens.

Now, a straight line is drawn passing through the origin and making an angle of $45^\circ$ with the x-axis. This line intersects our experimental curve at a specific point, $P$. Because the line makes a $45^\circ$ angle, its slope is $	an(45^\circ) = 1$. This gives us the equation of the line: $v = u$.

### The Master Equation

To find the coordinates of point $P$, we need to use the lens formula. The standard lens formula is $\frac{1}{v} - \frac{1}{u_{obj}} = \frac{1}{f}$.

However, we must be careful with our sign convention. Since the object is real and placed in front of the lens, the actual object distance is $-u$. The image is formed on a screen, meaning it is a real image, so the image distance is $+v$.

Substituting these into our formula, we get $\frac{1}{v} - \frac{1}{-u} = \frac{1}{f}$, which simplifies beautifully to $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. This is our master equation for the magnitudes of the distances.

### Final Calculation

At point $P$, the straight line and the curve intersect. This means the condition $v = u$ must satisfy our master equation.

Let's substitute $v = u$ into the equation: $\frac{1}{u} + \frac{1}{u} = \frac{1}{f}$.

Adding the terms on the left gives us $\frac{2}{u} = \frac{1}{f}$.

Solving for $u$, we find that $u = 2f$. Since we already established that $v = u$ at point $P$, it naturally follows that $v = 2f$ as well.

Therefore, the coordinates of point $P$ are **$(2f, 2f)$**.

### Physical Significance

What does this mathematical result actually mean in the real world? It corresponds to a very specific and important case in optics.

When an object is placed exactly at the center of curvature of a convex lens (which is at a distance of $2f$), the lens forms a real, inverted image on the exact opposite side, also at a distance of $2f$.

In this scenario, the size of the image is exactly equal to the size of the object, giving a magnification of $m = -1$. This is a classic setup often used in labs to quickly estimate the focal length of a lens!

Select Answer:

Visualized Solution

$v - u Graph Analysis$

$Equation of the Line$

$Lens Formula$

$Intersection Point P$

$Solving for u and v$

$Coordinates of P$

$Physical Significance$

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Final Calculation

Physical Significance

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v−u Graph Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the Line

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lens Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Intersection Point P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for u and v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinates of P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Significance