Question

A point object is kept in front of a convex spherical surface of radius of curvature $R$. Draw the ray diagram to show the formation of image and derive the relation between the object and image distance ($u$ and $v$) in terms of refractive index $n$ of the medium and $R$.

Hint:

When deriving spherical refraction formulas, always remember the exterior angle theorem ($\text{ext angle} = \text{sum of interior opposite angles}$) for the triangles formed by the incident/refracted ray, the normal, and the principal axis. It links physical optics to simple geometry instantly!

Answer 1

Concept: When light passes from one optical medium to another across a spherical boundary, refraction occurs according to Snell's Law. For a convex spherical surface separating two media of refractive indices $n_1$ and $n_2$ (where the light travels from medium 1 to medium 2), we can establish a geometric relationship between the object distance ($u$), image distance ($v$), and the radius of curvature ($R$). By considering paraxial rays (rays that make small angles with the principal axis), the trigonometric functions of the angles can be closely approximated by the angles themselves in radians ($\sin \theta \approx \theta$ and $\tan \theta \approx \theta$). This allows us to use simple geometric theorems, such as the exterior angle theorem of triangles, to derive the standard spherical refraction formula: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

Step 1: Ray Diagram Description and Construction Let us consider a convex spherical refracting surface separating a rarer medium of refractive index $n_1$ and a denser medium of refractive index $n_2$. Let $P$ be the pole, $C$ be the center of curvature, and $PC$ be the principal axis of this refracting surface. Let $R$ be the radius of curvature ($PC = +R$). A point object $O$ is placed on the principal axis in the rarer medium at a distance $u$ from the pole ($PO = -u$). A ray of light starting from $O$ is incident normally on the surface at $P$ and passes through undeviated along the principal axis. Another paraxial ray $OA$ is incident on the surface at point $A$ making an angle of incidence $i$ with the normal $AN$ (where $NC$ is the normal drawn through the center of curvature $C$). According to Snell's law, since the ray travels from a rarer to a denser medium ($n_2 > n_1$), it bends towards the normal along the direction $AI$. The two refracted rays meet at point $I$ on the principal axis, forming a real image $I$ at a distance $v$ from the pole ($PI = +v$). Let $\alpha$, $\beta$, and $\gamma$ be the small angles made by the incident ray $OA$, the refracted ray $AI$, and the normal $AC$ with the principal axis respectively. Let us drop a perpendicular $AM$ from point $A$ onto the principal axis.

Step 2: Geometrical Angle Relationships In triangle $\Delta OAC$, the exterior angle is equal to the sum of the two interior opposite angles. Therefore, for the incident ray: \[ i = \alpha + \gamma \quad \cdots (1) \] Similarly, in triangle $\Delta IAC$, the angle $\gamma$ is the exterior angle. Therefore: \[ \gamma = r + \beta \quad \Rightarrow \quad r = \gamma - \beta \quad \cdots (2) \] Since we are considering paraxial rays, the point $A$ lies very close to the pole $P$. Consequently, the angles $\alpha$, $\beta$, and $\gamma$ are extremely small. For very small angles expressed in radians, we can approximate the angles by their tangents: \[ \alpha \approx \tan \alpha = \frac{AM}{OM} \approx \frac{AM}{OP} \] \[ \beta \approx \tan \beta = \frac{AM}{MI} \approx \frac{AM}{PI} \] \[ \gamma \approx \tan \gamma = \frac{AM}{MC} \approx \frac{AM}{PC} \] Because point $A$ is very close to $P$, the point $M$ virtually coincides with the pole $P$, so $OM \approx OP$, $MI \approx PI$, and $MC \approx PC$.

Step 3: Application of Snell's Law According to Snell's Law of refraction at the interface: \[ n_1 \sin i = n_2 \sin r \] Since the angles $i$ and $r$ are very small for paraxial rays, we can approximate $\sin i \approx i$ and $\sin r \approx r$. Substituting these into Snell's law yields: \[ n_1 i = n_2 r \quad \cdots (3) \] Now, substitute the values of $i$ and $r$ from equations (1) and (2) into equation (3): \[ n_1 (\alpha + \gamma) = n_2 (\gamma - \beta) \] Expanding the brackets: \[ n_1 \alpha + n_1 \gamma = n_2 \gamma - n_2 \beta \] Rearranging the terms to group $\gamma$ on one side: \[ n_1 \alpha + n_2 \beta = (n_2 - n_1) \gamma \quad \cdots (4) \]

Step 4: Substituting Tangent Approximations and Applying Cartesian Sign Convention Substitute the fractional expressions for $\alpha$, $\beta$, and $\gamma$ into equation (4): \[ n_1 \left( \frac{AM}{OP} \right) + n_2 \left( \frac{AM}{PI} \right) = (n_2 - n_1) \left( \frac{AM}{PC} \right) \] Since $AM$ is common to all terms, we can divide the entire equation by $AM$: \[ \frac{n_1}{OP} + \frac{n_2}{PI} = \frac{n_2 - n_1}{PC} \quad \cdots (5) \] Now, applying the standard New Cartesian Sign Convention:
• Distance of the object, $OP = -u$ (measured against the direction of incident light)
• Distance of the image, $PI = +v$ (measured in the direction of incident light)
• Radius of curvature, $PC = +R$ (measured in the direction of incident light) Substituting these structural values into equation (5): \[ \frac{n_1}{-u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R} \] Rearranging the terms cleanly gives the final required relationship: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \] If the first medium is air or vacuum ($n_1 = 1$) and the second medium has a refractive index $n_2 = n$, the relation simplifies directly to: \[ \frac{n}{v} - \frac{1}{u} = \frac{n - 1}{R} \]