The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m

Q: Why did we take $u$ as negative but $v$ as positive?

According to the Cartesian sign convention, the direction of incident light is taken as positive. Since the object is placed in front of the surface (against the light direction), $u$ is negative. The real image forms behind the surface (in the direction of light), so $v$ is positive.

Q: Why does the wavelength change when light enters the medium, but not the frequency?

Frequency is a fundamental property of the light source and represents the energy of the photons, which cannot change. When light enters a denser medium, it slows down. Since $v = f\lambda$, a decrease in speed $v$ must result in a proportional decrease in wavelength $\lambda$.

Q: How did we know that $n_2 = \frac{3}{2} n_1$?

The refractive index of a medium is defined as the ratio of the speed of light in vacuum to its speed in the medium ($n = c/v$). Since $v = f\lambda$ and $f$ is constant, $n$ is inversely proportional to $\lambda$. If $\lambda_2 = \frac{2}{3}\lambda_1$, then $n_2 = \frac{3}{2}n_1$.

Q: Can we just assume the refractive index of air $n_1$ is 1?

Yes, for all practical purposes in JEE problems, the refractive index of air is taken as 1. However, as you saw in the calculation, $n_1$ cancels out completely from both sides of the equation, so its exact value doesn't even matter!

Q: What would happen if the image formed was virtual?

If the image were virtual, it would form on the same side as the object. In that case, the image distance $v$ would be taken as negative according to the sign convention, which would completely change the calculation for $R$.

Question

The image of an object placed in air formed by a convex refracting surface is at a distance of $10	ext{ m}$ behind the surface. The image is real and is at $\frac{2}{3}$ of the distance of the object from the surface. The wavelength of light inside the surface is $\frac{2}{3}$ times the wavelength in air. The radius of the curved surface is $\frac{x}{13}	ext{ m}$. The value of $x$ is ............... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a convex refracting surface that acts as a boundary between two media: air and a denser transparent material. We are told that an object is placed in the air, and its real image is formed inside the denser medium.

The problem gives us a crucial piece of geometric information: the image is formed at a distance of $10	ext{ m}$ behind the surface. Furthermore, this image distance is exactly $\frac{2}{3}$ of the object's distance from the surface.

Let's translate this into mathematical constraints. If the image distance is $v = 10	ext{ m}$, and $v = \frac{2}{3} |u|$, we can easily find the object distance:
$$10 = \frac{2}{3} |u| \implies |u| = 15	ext{ m}$$

By standard sign convention, since the object is placed in front of the refracting surface, we take $u = -15	ext{ m}$.

### The Refractive Index Connection

Next, we need to determine the refractive index of the denser medium. The problem states that the wavelength of light inside the surface is $\frac{2}{3}$ times its wavelength in air.

We know from wave optics that the refractive index of a medium is inversely proportional to the wavelength of light in that medium ($n \propto \frac{1}{\lambda}$). Therefore, if the wavelength decreases by a factor of $\frac{2}{3}$, the refractive index must increase by the reciprocal factor.

Assuming the refractive index of air is $n_1$, the refractive index of the medium $n_2$ will be:
$$n_2 = \frac{3}{2} n_1$$

### The Master Equation

Now we have all the pieces required to use the formula for refraction at a single spherical surface:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

Let's substitute our known values into this master equation:
$$\frac{\frac{3}{2} n_1}{10} - \frac{n_1}{-15} = \frac{\frac{3}{2} n_1 - n_1}{R}$$

Notice how $n_1$ appears in every single term? This is a beautiful moment in physics where the absolute refractive index doesn't matter, only the relative ratio does. We can cancel $n_1$ out completely:
$$\frac{3}{20} + \frac{1}{15} = \frac{\frac{1}{2}}{R}$$

### Final Calculation

All that remains is some basic fraction addition. Let's find a common denominator for $20$ and $15$, which is $60$:
$$\frac{9}{60} + \frac{4}{60} = \frac{1}{2R}$$
$$\frac{13}{60} = \frac{1}{2R}$$

Cross-multiplying to solve for the radius of curvature $R$:
$$2R = \frac{60}{13} \implies R = \frac{30}{13}	ext{ m}$$

The problem states that the radius of the curved surface is $\frac{x}{13}	ext{ m}$. By directly comparing our result with the given expression, we find:
**$x = 30$**

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Finding Object Distance $u$

Refractive Index Relation

Spherical Refraction Formula

Simplifying the Equation

Calculating Radius $R$

Conceptual Reflection

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

The Refractive Index Connection

The Master Equation

Final Calculation

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