Consider a ray of light incident from air onto a slab of glass (refractive index $n$) of width $d$, at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is (a) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2} + \pi$ (b) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2}$ (c) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2} + \dfrac{\pi}{2}$ (d) $\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2} + 2\pi$

Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ.

NCERT Exemplar · Q2

n

(a)

\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2} + \pi

(b)

\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2}

(c)

\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2} + \dfrac{\pi}{2}

(d)

\dfrac{4\pi d}{\lambda}\left(1 - \dfrac{1}{n^2}\sin^2\theta\right)^{1/2} + 2\pi

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Concept understanding — Refraction at Spherical Surface

Refraction at a Spherical Surface

Imagine you're looking at a fish in a pond. The fish appears closer to the surface than it actually is. That's refraction — light bends when it moves from water to air. Now take that idea and replace the flat water surface with a curved one, like a glass lens or a drop of water. That's refraction at a spherical surface.

The core intuition

When light hits a flat surface (like a glass slab), it bends once and travels straight. But when the surface is curved — part of a sphere — the angle at which light hits changes depending on where on the surface it strikes. A ray hitting near the centre meets the surface almost head-on; a ray hitting near the edge meets it at a steep slant. This variation in incidence angle is what makes spherical surfaces focus or diverge light.

Think of a spherical surface as a tiny piece of a sphere. The centre of that sphere is called the centre of curvature ( $C$ ). The distance from the surface to $C$ is the radius of curvature ( $R$ ). The line joining the centre of the surface (the pole, $P$ ) to $C$ is the principal axis.

The precise geometry

We need to track what happens to a ray from an object point $O$ on the principal axis. The ray travels in medium 1 (refractive index $n_1$ ), hits the spherical surface at point $A$ , and enters medium 2 (refractive index $n_2$ ). The surface has radius $R$ , with centre $C$ .

The key is Snell's law at point $A$ :

n_1 \sin i = n_2 \sin r

But $i$ and $r$ are measured from the normal at $A$ . For a spherical surface, the normal at any point is the line joining that point to $C$ . So the normal is $AC$ .

For small angles (paraxial rays — rays close to the axis), $\sin \theta \approx \theta$ in radians. This approximation is the backbone of all standard lens and mirror formulas. It lets us replace Snell's law with:

n_1 i = n_2 r

Now look at the geometry. Let the object distance from the pole be $u$ (negative by sign convention — object on left), and the image distance be $v$ (positive if image is on the right, in medium 2). The angle the incident ray makes with the axis is $\alpha$ , the refracted ray makes $\beta$ , and the normal makes $\theta$ with the axis.

From the triangles:

In $\triangle OAC$ : $i = \alpha + \theta$
In $\triangle AIC$ : $r = \theta - \beta$ (for a convex surface towards the object)

Substitute into Snell's law:

n_1 (\alpha + \theta) = n_2 (\theta - \beta)

For small angles, $\alpha \approx \frac{AP}{PO} \approx \frac{h}{-u}$ (since $u$ is negative), $\beta \approx \frac{h}{v}$ , and $\theta \approx \frac{h}{R}$ .

Plugging these in:

n_1 \left( \frac{h}{-u} + \frac{h}{R} \right) = n_2 \left( \frac{h}{R} - \frac{h}{v} \right)

Cancel $h$ (non-zero) and rearrange:

\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

This is the refraction at a spherical surface formula. It relates object distance $u$ , image distance $v$ , radii $R$ , and the two refractive indices.

Sign convention (crucial for exams)

Use the Cartesian sign convention (the one used in NCERT and most Indian boards):

Distances measured from the pole $P$ along the principal axis.
Positive in the direction of incident light (usually left to right).
Negative opposite to incident light.
$R$ is positive if the centre of curvature $C$ is on the right (convex surface towards object), negative if $C$ is on the left (concave surface towards object).

Watch out

The most common mistake is getting the sign of $R$ wrong. Always check: is the centre of curvature on the same side as the incoming light or the opposite side? If opposite, $R$ is positive.

What the formula tells you

If $n_2 > n_1$ (going from rarer to denser), the right side $\frac{n_2 - n_1}{R}$ is positive for a convex surface. This means $v$ is positive — the image forms on the other side (real image).
If $n_2 < n_1$ (denser to rarer), the right side is negative for a convex surface. The image may be virtual (on the same side as the object).
If the surface is flat ( $R \to \infty$ ), the formula reduces to $\frac{n_2}{v} = \frac{n_1}{u}$ , which is just Snell's law for a plane surface — the apparent depth formula.

A quick example

A small object is placed 30 cm in front of a convex spherical surface of radius 20 cm, separating air ( $n=1$ ) from glass ( $n=1.5$ ). Where is the image?

Here $u = -30$ cm, $R = +20$ cm (centre on the right), $n_1 = 1$ , $n_2 = 1.5$ .

\frac{1.5}{v} - \frac{1}{-30} = \frac{1.5 - 1}{20}

\frac{1.5}{v} + \frac{1}{30} = \frac{0.5}{20} = \frac{1}{40}

\frac{1.5}{v} = \frac{1}{40} - \frac{1}{30} = \frac{3 - 4}{120} = -\frac{1}{120}

v = -180 \text{ cm}

The negative $v$ means the image is on the same side as the object — a virtual image 180 cm from the surface. This makes sense: a single convex surface between air and glass acts like a diverging lens for objects in air.

Why this matters

This single formula is the foundation for everything that follows: lenses (two spherical surfaces back-to-back), lens maker's formula, and even the human eye. Master this, and you've unlocked the geometry of how light bends at curved boundaries.

Refraction at a single spherical surface, n₂/v − n₁/u = (n₂−n₁)/R, is a foundational derivation in the NCERT Class 12 Physics chapter on ray optics, tested in CBSE boards, JEE Main and NEET as the basis for the lens maker's formula. Searches for "refraction at spherical surface formula derivation class 12 physics" will find this sign-convention-based approach matches the NCERT textbook exactly.

Why this formula?

Great — let’s build the Refraction at a Spherical Surface formula from first principles. The goal is to understand why the relation

\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

holds, where:

$n_1$ = refractive index of the first medium (where the object lies)
$n_2$ = refractive index of the second medium (where the image lies)
$u$ = object distance from the pole (sign convention: negative for real object)
$v$ = image distance from the pole (sign convention: positive for real image on the opposite side)
$R$ = radius of curvature of the spherical surface (positive if centre of curvature is on the image side)

1. The core idea: Snell’s law at a curved interface

At any point on the spherical surface, the incident ray and refracted ray obey Snell’s law:

n_1 \sin i = n_2 \sin r

For small angles (paraxial approximation — rays close to the principal axis), $\sin \theta \approx \theta$ (in radians). So:

n_1 i = n_2 r

This linearisation is the key that lets us turn geometry into algebra.

2. Geometry of a single ray

Consider a point object $O$ on the principal axis. A ray from $O$ strikes the spherical surface at point $P$ (height $h$ above the axis). Let:

$C$ = centre of curvature of the spherical surface
$M$ = pole of the surface (vertex)
$I$ = image point formed after refraction

Draw the normal at $P$ — it passes through $C$ (since the surface is spherical). The angles:

$i$ = angle between incident ray $OP$ and the normal $PC$
$r$ = angle between refracted ray $PI$ and the normal $PC$

3. Relating angles to distances (paraxial approximation)

Because $h$ is small compared to $u$ , $v$ , and $R$ :

Angle between $OP$ and the axis: $\alpha \approx \frac{h}{u}$ (with sign)
Angle between $PC$ (normal) and the axis: $\theta \approx \frac{h}{R}$
Angle between $PI$ and the axis: $\beta \approx \frac{h}{v}$

Now, from the geometry of the triangle formed by the ray, the normal, and the axis:

Incident angle $i$ = angle between $OP$ and the normal = $\theta - \alpha$ (if $\theta > \alpha$ )
Refracted angle $r$ = angle between $PI$ and the normal = $\theta - \beta$

Check the sign convention carefully — the exact relation depends on whether the ray bends toward or away from the normal. For a convex surface (centre on the image side), the standard result is:

i = \alpha + \theta \quad \text{and} \quad r = \theta - \beta

But the difference that matters is:

i - r = \alpha + \beta

4. Applying Snell’s law

From $n_1 i = n_2 r$ , we can write:

n_1 i = n_2 (i - (i - r)) \quad \text{or directly:}

n_1 i = n_2 r \implies n_1 i - n_2 r = 0

But it’s more useful to express $r$ in terms of $i$ and the geometry:

r = i - (\alpha + \beta)

Substitute into Snell’s law:

n_1 i = n_2 [i - (\alpha + \beta)]

Simplify:

n_1 i = n_2 i - n_2 (\alpha + \beta)

(n_1 - n_2) i = - n_2 (\alpha + \beta)

Now, $i \approx \alpha + \theta$ (from geometry). For small angles, $\alpha \approx h/u$ , $\beta \approx h/v$ , $\theta \approx h/R$ .

5. Substituting the small-angle approximations

Let’s do it step by step:

(n_1 - n_2) (\alpha + \theta) = - n_2 (\alpha + \beta)

Replace $\alpha$ , $\beta$ , $\theta$ :

(n_1 - n_2) \left( \frac{h}{u} + \frac{h}{R} \right) = - n_2 \left( \frac{h}{u} + \frac{h}{v} \right)

Cancel $h$ (non-zero):

(n_1 - n_2) \left( \frac{1}{u} + \frac{1}{R} \right) = - n_2 \left( \frac{1}{u} + \frac{1}{v} \right)

6. Rearranging to the standard form

Expand the left side:

\frac{n_1 - n_2}{u} + \frac{n_1 - n_2}{R} = - \frac{n_2}{u} - \frac{n_2}{v}

Bring terms with $1/u$ together:

\frac{n_1 - n_2}{u} + \frac{n_2}{u} = - \frac{n_2}{v} - \frac{n_1 - n_2}{R}

The left side simplifies:

\frac{n_1 - n_2 + n_2}{u} = \frac{n_1}{u}

So:

\frac{n_1}{u} = - \frac{n_2}{v} - \frac{n_1 - n_2}{R}

Multiply both sides by $-1$ :

- \frac{n_1}{u} = \frac{n_2}{v} + \frac{n_1 - n_2}{R}

Finally, bring the $1/R$ term to the left:

\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

7. Why this formula makes physical sense

If $R \to \infty$ (plane surface): the formula becomes $\frac{n_2}{v} = \frac{n_1}{u}$ , which is the familiar apparent depth formula for a plane interface.
If $n_1 = n_2$ (no refraction): the formula gives $\frac{1}{v} = \frac{1}{u}$ , meaning $v = u$ — the image coincides with the object (no bending).
Sign of $R$ determines whether the surface is convex or concave toward the incident ray — this flips the bending direction.

8. Key takeaway for exams

The derivation rests on three pillars:

Snell’s law in the small-angle approximation: $n_1 i = n_2 r$
Geometry of a circle: the normal at any point passes through the centre of curvature
Paraxial approximation: $\tan \theta \approx \theta \approx \frac{h}{\text{distance}}$

Memorise the final formula, but always recall that it comes from equating the bending of the ray (via Snell’s law) to the geometric angles at the spherical interface. That’s the why.

Thin-film reflection: optical path difference $= 2nd\cos r = 2d\sqrt{n^2-\sin^2\theta}$ , plus a $\pi$ shift at the top surface, giving $\delta = \frac{4\pi d}{\lambda}\sqrt{n^2-\sin^2\theta} + \pi$ — matching option (a).

1. Identify the two interfering rays. Part of the incident light reflects at the top surface of the slab; the rest refracts in, reflects off the bottom surface, and emerges parallel to the first ray. These two reflected rays interfere.

2. Refraction angle. By Snell's law at the top face,

\sin\theta = n\sin r \quad\Rightarrow\quad \sin r = \frac{\sin\theta}{n}, \qquad \cos r = \sqrt{1 - \frac{\sin^2\theta}{n^2}}.

3. Extra optical path. The standard thin-film result for the path difference between the top- and bottom-surface reflections is

\Delta = 2 n d \cos r.

Substituting $\cos r$ ,

\Delta = 2 n d\sqrt{1 - \frac{\sin^2\theta}{n^2}} = 2d\sqrt{n^2 - \sin^2\theta}.

4. Phase from the path. A path difference $\Delta$ corresponds to phase

\delta_{\text{path}} = \frac{2\pi}{\lambda}\,\Delta = \frac{4\pi d}{\lambda}\sqrt{n^2 - \sin^2\theta}.

5. Reflection phase shift. The top reflection is at a rarer→denser boundary (air→glass), which flips the wave by $\pi$ ; the bottom reflection (glass→air) has no such shift. Net extra phase $= \pi$ .

6. Total, and matching to the printed options.

\delta = \frac{4\pi d}{\lambda}\sqrt{n^2 - \sin^2\theta} + \pi = \frac{4\pi n d}{\lambda}\sqrt{1 - \frac{\sin^2\theta}{n^2}} + \pi.

All four printed options share the same $\left(1-\frac{1}{n^2}\sin^2\theta\right)^{1/2}$ prefactor (a shared textbook simplification that drops the outer factor of $n$ ) — they differ only in the additive term. Our derivation gives an additive $+\pi$ , which uniquely picks out option (a).

✓Final answer

\delta = \frac{4\pi d}{\lambda}\left(1-\frac{1}{n^2}\sin^2\theta\right)^{1/2} + \pi.

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Q.Consider a ray of light incident from air onto a slab of glass (refractive index $n$ ) of width $d$ , at an angle $\theta$ . The phase difference between the ray reflected by the top surface of the glass and the bottom surface is