Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvatu

Q: Why is the radius of curvature for the first surface negative?

Because the surface is convex towards the outside (air), but from the perspective of the light rays originating *inside* the glass rod, the boundary curves inward towards them. Thus, the center of curvature lies against the direction of incident light, making it negative according to the Cartesian sign convention.

Q: Why does the intermediate image $I_1$ act as a real object for the second surface?

The rays from the first surface converge and actually cross at $I_1$ before reaching the second surface. Since the rays physically intersect and then diverge towards the second surface, $I_1$ serves as a real object for it.

Q: What does it mean when rays become parallel to the axis?

It means that the image is formed at infinity. In the refraction formula, we set the image distance $v$ to infinity, which makes the term $\frac{\mu}{v}$ equal to zero.

Q: How do we determine the object distance $u_2$ for the second surface?

The intermediate image $I_1$ is formed at a distance $v_1$ from the first surface. The total distance between the surfaces is $d$. Since $I_1$ is between the surfaces, its distance to the second surface is $d - v_1$. Applying the sign convention (light travels left to right), $u_2 = -(d - v_1)$.

Q: Could we solve this using the principle of reversibility?

Absolutely! If you imagine light rays traveling backwards (from right to left) parallel to the axis inside $S_2$, they would refract at the convex surface and converge at its first focal point. Calculating this focal length gives 20 cm, meaning the object $I_1$ must be 20 cm in front of $S_2$.

Question

Two identical glass rods $S_1$ and $S_2$ (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance $d$ as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light $P$ is placed inside rod $S_1$ on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside $S_2$. The distance $d$ is

Options:

(A) 60 cm

(B) 70 cm

(C) 80 cm

(D) 90 cm

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with a fascinating optics problem involving two identical glass rods, $S_1$ and $S_2$, separated by an air gap of distance $d$. Both rods have a refractive index of $1.5$ and feature a convex end with a radius of curvature of $10	ext{ cm}$.

A point source $P$ is placed inside the first rod, $S_1$, exactly $50	ext{ cm}$ away from its curved boundary. Our goal is to find the separation distance $d$ such that the light rays, after passing through the air gap and entering the second rod $S_2$, become perfectly parallel to the principal axis.

### The First Refraction: Escaping the Glass

To trace the journey of the light rays, we must analyze the refraction at each spherical boundary individually. The master equation governing refraction at a single spherical surface is:

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

Let's focus on the first boundary, where light exits $S_1$ and enters the air gap. The light originates in the glass, so our initial medium has $\mu_1 = 1.5$, and it enters the air, so $\mu_2 = 1$.

Applying the Cartesian sign convention, the object distance $u_1$ is $-50	ext{ cm}$. Now, pay close attention to the radius of curvature. Although the rod's end is convex towards the outside, from the perspective of the light rays traveling *inside* the rod, the boundary curves inward. Therefore, the center of curvature lies against the direction of incident light, making $R_1 = -10	ext{ cm}$.

Substituting these values into our master equation:

$$\frac{1}{v_1} - \frac{1.5}{-50} = \frac{1 - 1.5}{-10}$$

$$\frac{1}{v_1} + \frac{3}{100} = \frac{0.5}{10} = \frac{5}{100}$$

$$\frac{1}{v_1} = \frac{2}{100} \implies v_1 = +50	ext{ cm}$$

This positive result tells us that the rays converge to form a real intermediate image, $I_1$, exactly $50	ext{ cm}$ to the right of the first surface.

### The Second Refraction: Entering the Second Rod

This intermediate image $I_1$ now acts as the object for the second refraction at the surface of rod $S_2$. The problem states a crucial condition: after entering $S_2$, the rays become parallel to the axis. In the language of optics, this means the final image is formed at infinity, so $v_2 = \infty$.

Let's set up the parameters for this second boundary. Light is traveling from air ($\mu_1 = 1$) into the glass rod ($\mu_2 = 1.5$). The surface of $S_2$ bulges out towards the incoming rays, meaning its center of curvature lies in the direction of light travel. Thus, $R_2 = +10	ext{ cm}$.

Applying the refraction formula once more to find the object distance $u_2$:

$$\frac{1.5}{\infty} - \frac{1}{u_2} = \frac{1.5 - 1}{+10}$$

Since any finite number divided by infinity approaches zero, the equation simplifies beautifully:

$$0 - \frac{1}{u_2} = \frac{0.5}{10} = \frac{1}{20}$$

$$u_2 = -20	ext{ cm}$$

This tells us that the object for the second surface (which is our intermediate image $I_1$) must be located $20	ext{ cm}$ to the left of the second surface.

### Final Calculation: Bridging the Gap

We now have all the pieces of the puzzle. The intermediate image $I_1$ is located $50	ext{ cm}$ to the right of the first surface, and it must simultaneously be $20	ext{ cm}$ to the left of the second surface.

Therefore, the total distance $d$ between the two curved surfaces is simply the sum of these two segments:

$$d = v_1 + |u_2|$$

$$d = 50	ext{ cm} + 20	ext{ cm} = \mathbf{70	ext{ cm}}$$

By systematically breaking down the complex optical system into two distinct refraction events, we've elegantly arrived at the solution. The distance $d$ must be $70	ext{ cm}$.

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Visualized Solution

$A na l y z in g t h e S e t u p$

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

$F i r s t R e f r a c t i o n (Gl a ss t o A i r)$

$C a l c u l a t in g I ma g e P os i t i o n I_{1}$

$I n t er m e d ia t e I ma g e I_{1}$

$S eco n d R e f r a c t i o n (A i r t o Gl a ss)$

$P a r am e t er s f or S eco n d S u r f a ce$

$C a l c u l a t in g O bj ec t D i s t an ce u_{2}$

$F in d in g t h e D i s t an ce d$

$P r in c i pl e o f R e v er s ibi l i t y$

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

The First Refraction

The Second Refraction

Final Calculation

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Refraction (Glass to Air)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Image Position I1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Intermediate Image I1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Refraction (Air to Glass)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parameters for Second Surface

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Object Distance u2​

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

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