A slab of material of refractive index 2 shown in figure has a curved surface APB of radius of curva

Q: Why do we ignore the plane surface CD and the water?

Because the observer is on the left. The light rays that reach the observer's eye travel from the object O, leftwards, and exit through the surface APB. They never travel rightwards to hit CD.

Q: Why is the object distance $u$ taken as negative?

If we place the origin at the pole P and take the direction of incident light (leftwards) as positive, then the object O, which is to the right of P, lies on the negative axis.

Q: Why is the radius of curvature $R$ negative?

The surface APB bulges towards the left. Its center of curvature lies to the right of P. Since the right side is the negative axis in our convention, $R$ is negative.

Q: What does a negative image distance $v$ mean?

A negative $v$ means the image is formed on the negative axis, which is to the right of P. Since the light has already passed into the air (to the left), an image formed on the right is virtual.

Q: Can I solve this by taking the rightward direction as positive?

Absolutely! If right is positive, then incident light travels in the negative direction. The formula remains the same, but you must be consistent. $u = +15$, $R = +10$, but the formula $\frac{\mu_2}{v} - \frac{\mu_1}{u}$ assumes light travels in the positive direction. It's safer to always take the direction of incident light as positive.

Question

A slab of material of refractive index 2 shown in figure has a curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in the figure. An object O is placed at a distance of 15 cm from the pole P as shown. The distance of the final image of O from P, as viewed from the left is ……

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are the observer standing to the left of the glass slab, looking into the curved surface $APB$. The light rays that carry the visual information of the object $O$ must travel from the object, move leftwards through the slab, and refract at the surface $APB$ to reach your eyes.

Notice that the plane surface $CD$ and the water on the far right are completely irrelevant to this specific observer. The rays reaching the observer's eyes never interact with the right side of the setup. Therefore, this problem simplifies to a single refraction at a spherical surface.

### Setting Up the Sign Convention

To solve this mathematically, we need a robust sign convention. Let's place our origin at the pole $P$ of the curved surface.

A standard and foolproof convention is to take the **direction of incident light as positive**. Since the light travels from the object $O$ towards the observer on the left, the **leftward direction is positive**.

Consequently, anything located to the right of the pole $P$ will have a negative coordinate:
- The object $O$ is $15 	ext{ cm}$ to the right, so the object distance $u = -15 	ext{ cm}$.
- The surface $APB$ bulges to the left, meaning its center of curvature $C$ lies to the right. Thus, the radius of curvature $R = -10 	ext{ cm}$.
- The light originates in the slab, so the initial refractive index $\mu_1 = 2.0$.
- The light enters the air, so the final refractive index $\mu_2 = 1.0$.

### The Master Equation

For refraction at a single spherical surface, the governing equation is:

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

Let's carefully substitute our parameters into this equation. Watch out for the negative signs, as they are the most common trap in optics problems!

$$ \frac{1.0}{v} - \frac{2.0}{-15} = \frac{1.0 - 2.0}{-10} $$

### Final Calculation

Now, it's just a matter of simple algebra. The two negative signs on the left side cancel out:

$$ \frac{1}{v} + \frac{2}{15} = \frac{-1.0}{-10} $$

$$ \frac{1}{v} + \frac{2}{15} = \frac{1}{10} $$

Isolating the $\frac{1}{v}$ term:

$$ \frac{1}{v} = \frac{1}{10} - \frac{2}{15} $$

Taking a common denominator of $30$:

$$ \frac{1}{v} = \frac{3 - 4}{30} = -\frac{1}{30} $$

Inverting both sides gives us the final image position:

$$ v = -30 	ext{ cm} $$

### Interpreting the Result

What does $v = -30 	ext{ cm}$ mean physically? Since our positive direction was defined as leftwards, a negative $v$ indicates that the image is formed $30 	ext{ cm}$ to the **right** of the pole $P$.

Because the refracted rays are diverging into the air on the left, they only *appear* to intersect when traced backwards to the right. This means the image is **virtual**. The final distance of this image from the pole $P$ is exactly **$30 	ext{ cm}$**.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Sign Convention \& Parameters}

\text{Spherical Refraction Formula}

\text{Substituting the Values}

\text{Solving for } v

\text{Final Calculation}

\text{Interpreting the Result}

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

Setting Up the Sign Convention

The Master Equation

Final Calculation

Interpreting the Result

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