Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I

Q: Why is the radius of curvature taken as negative?

According to the Cartesian sign convention, the direction of incident light is taken as positive. Since the light travels from Region I to Region II (left to right), the right side is positive. The surface is concave towards Region I, meaning its center of curvature lies to the left of the pole. Therefore, the radius is negative.

Q: What does a negative image distance mean physically?

A negative image distance means the image is formed on the same side of the refracting surface as the object (in Region I). This indicates that the refracted rays diverge, and the image formed is virtual.

Q: Why did we use $\mu_1 = 1.25$ and not $1.4$?

In the spherical refraction formula, $\mu_1$ always represents the refractive index of the medium where the incident rays originate. Since the object is in Region I, the incident rays start there, making $\mu_1 = 1.25$.

Q: Can we just use the standard lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ here?

No, the standard lens formula applies to thin lenses with two refracting surfaces in the same surrounding medium. Here, we are dealing with a single spherical refracting surface separating two different media, so we must use the specific formula for it.

Q: What if the object was placed in Region II instead?

If the object were in Region II, the incident light would travel from right to left. You would need to reverse your sign convention (left becomes positive) or keep right as positive but make $u$ positive. Also, $\mu_1$ would become $1.4$ and $\mu_2$ would be $1.25$.

Question

Region I and II are separated by a spherical surface of radius $25	ext{ cm}$. An object is kept in region I at a distance of $40	ext{ cm}$ from the surface. The distance of the image from the surface is

Options:

(A) $55.44	ext{ cm}$

(B) $9.52	ext{ cm}$

(C) $18.23	ext{ cm}$

(D) $37.58	ext{ cm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in Region I, looking towards Region II through a curved glass window. This window is a spherical refracting surface. The object is placed in Region I, and we need to find exactly where its image forms.

To do this, we must carefully apply the Cartesian sign convention. Since light travels from left to right, we take the right direction as positive.

The object is placed $40	ext{ cm}$ to the left of the surface, so the object distance is $u = -40	ext{ cm}$. The surface is concave towards Region I, meaning its center of curvature also lies to the left. Thus, the radius of curvature is $R = -25	ext{ cm}$.

### The Master Equation

The behavior of light refracting at a single spherical surface is governed by a beautiful relationship:

$$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $$

Here, $\mu_1$ is the refractive index of the medium where the incident light originates (Region I, so $\mu_1 = 1.25$), and $\mu_2$ is the refractive index of the medium the light enters (Region II, so $\mu_2 = 1.4$).

### Executing the Calculation

Let's substitute our known values into the master equation:

$$ \frac{1.4}{v} - \frac{1.25}{-40} = \frac{1.4 - 1.25}{-25} $$

First, we simplify the right side of the equation. The difference in refractive indices is $1.4 - 1.25 = 0.15$. Dividing this by $-25$ gives us $-\frac{3}{500}$.

On the left side, the two negative signs cancel out, leaving us with $+\frac{1.25}{40}$, which simplifies to $\frac{1}{32}$.

Now, we isolate the term containing our unknown image distance, $v$:

$$ \frac{1.4}{v} = -\frac{3}{500} - \frac{1}{32} $$

To subtract these fractions, we find a common denominator, which is $4000$.

$$ \frac{1.4}{v} = -\frac{24}{4000} - \frac{125}{4000} = -\frac{149}{4000} $$

### Final Conclusion

Finally, we solve for $v$:

$$ v = -\frac{1.4 	imes 4000}{149} = -\frac{5600}{149} \approx -37.58	ext{ cm} $$

The negative sign is crucial here. It tells us that the image is formed on the **same side as the object** (Region I). This means the refracted rays diverge, and the image is **virtual**. The distance of this image from the surface is $37.58	ext{ cm}$.

Animated Solution for Physics - Optics: Region I and II are separated by a spherical surface of radius $25 cm$ . An object is kept in region I at a distance of $40 cm$ from the surface. The distance of the image from the surface is

Select Answer:

Visualized Solution

Visualizing the Setup

Formula for Spherical Refraction

Applying Sign Convention

Simplifying the Equation

Isolating the Unknown

Final Calculation

Conclusion

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

The Master Equation

Executing the Calculation

Final Conclusion

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