Figure shows an irregular block of material of refractive index \sqrt{2}. A ray of light strikes the

Q: Why is the object distance $u$ taken as $-\infty$ for the second surface?

Because the ray inside the block becomes perfectly horizontal after the first refraction, meaning it is parallel to the principal axis $PQ$. In optics, parallel rays are considered to originate from an object at infinity.

Q: How do we know the radius of curvature $R$ is positive?

The light travels from left to right. The spherical surface $CD$ is convex towards the incident ray (it bulges into the block), so its center of curvature lies to the right of the pole $O$. According to the standard Cartesian sign convention, distances measured in the direction of incident light are positive, making $R = +0.4\text{ m}$.

Q: What if the ray wasn't parallel to the principal axis after the first refraction?

You would need to trace the ray backwards to find the exact virtual object position created by the first refraction. That finite distance would then be used as $u$ in the spherical refraction formula.

Q: Why did we approximate $\sqrt{2}$ as $1.414$?

The problem gives the second refractive index as $1.514$, which is a very specific number. This is a strong hint from the examiner that $\sqrt{2}$ should be taken as $1.414$ to make the subtraction $1.514 - 1.414$ yield a clean and easy-to-calculate $0.100$.

Q: Does the height at which the ray strikes the spherical surface $CD$ matter?

In paraxial optics, which this problem assumes, all rays parallel to the principal axis will converge at the exact same focal point on the axis, regardless of their height above the axis.

Question

Figure shows an irregular block of material of refractive index $\sqrt{2}$. A ray of light strikes the face $AB$ as shown in the figure. After refraction it is incident on a spherical surface $CD$ of radius of curvature $0.4	ext{ m}$ and enters a medium of refractive index $1.514$ to meet $PQ$ at $E$. Find the distance $OE$ upto two places of decimal.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's embark on this fascinating journey through optics. We are presented with an irregular block of material with a refractive index of $\mu = \sqrt{2}$. A ray of light strikes the inclined face $AB$. The first crucial step is to understand the geometry of this interface.

The face $AB$ is tilted at an angle of $60^\circ$ to the horizontal. Consequently, the normal to this face will be tilted at $30^\circ$ to the horizontal. The problem states that the ray strikes this face at an angle of incidence $i = 45^\circ$.

### The First Refraction: Snell's Law

To determine the path of the ray inside the block, we apply Snell's Law at the face $AB$. The light originates from air ($\mu_1 = 1$) and enters the block ($\mu_2 = \sqrt{2}$).

$$1 \cdot \sin 45^\circ = \sqrt{2} \cdot \sin r$$

Substituting the value of $\sin 45^\circ$, we get:

$$\frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin r$$

$$\sin r = \frac{1}{2} \implies r = 30^\circ$$

Here lies the beautiful geometric trick of this problem! The normal is at $30^\circ$ to the horizontal, and the ray bends by exactly $30^\circ$ from the normal. This means the refracted ray becomes perfectly horizontal, traveling parallel to the principal axis $PQ$.

### The Second Refraction: Spherical Surface

Next, this horizontal ray encounters the spherical surface $CD$. Because the ray is parallel to the principal axis, it effectively comes from an object at infinity, so $u = -\infty$. We use the spherical refraction formula to find where it converges on the axis.

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

Let's plug in our values. The ray is coming from the block, so $\mu_1 = \sqrt{2} \approx 1.414$. It enters the new medium with $\mu_2 = 1.514$. The radius of curvature $R$ is positive $0.4	ext{ m}$ because the surface is convex towards the incident ray.

$$\frac{1.514}{v} - 0 = \frac{1.514 - 1.414}{0.4}$$

### Final Calculation

Notice how elegantly the numbers are designed to cancel out. The difference in the refractive indices is exactly $0.100$.

$$\frac{1.514}{v} = \frac{0.100}{0.4} = \frac{1}{4}$$

Multiplying $1.514$ by $4$, we obtain our final image distance, $v$:

$$v = 1.514 	imes 4 = 6.056	ext{ m}$$

Rounding to two decimal places, the distance $OE$ is **$6.06	ext{ m}$**.

Enter Numerical Value:

Visualized Solution

$G eo m e t r y o f F a ce A B$

$S n e l l^{'} s L a w a t A B$

$A n g l e o f R e f r a c t i o n$

$P a t h I n s i d e t h e B l oc k$

$R e f r a c t i o n a t S p h er i c a l S u r f a ce$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g I ma g e D i s t an ce$

$F ina l A n s w er$

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

The First Refraction

The Second Refraction

Final Calculation

Similar Questions

A quarter cylinder of radius $R$ and refractive index $1.5$ is placed on a table. A point object $P$ is kept at a distance of $m R$ from it. Find the value of $m$ for which a ray from $P$ will emerge parallel to the table as shown in figure.

In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is $R$ . Determine the focal length of this system.

A point object is placed at the centre of a glass sphere of radius $6 cm$ and refractive index $1.5$ . The distance of the virtual image from the surface of the sphere is

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Snell′s Law at AB

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Inside the Block

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Refraction at Spherical Surface

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

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

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