Derive the following expression for the refraction at concave spherical surface:

[latex]\frac{μ}{v}−\frac{1}{u}=\frac{μ−1}{R}[/latex]

Hint:A spherical mirror is a section of a sphere with a reflective surface. If the inner surface is reflective, the mirror is concave; if the outer surface reflects, it is convex. In this case, we are working with a concave mirror and need to derive the related expression.

Solution:

Consider a concave mirror, represented by [latex]MPN[/latex], with [latex]mu[/latex] as the refractive index of the medium surrounding the mirror. Here:

– [latex]P[/latex] is the mirror’s pole,

– [latex]O[/latex] is the center of curvature, and

– [latex]PC[/latex] is the principal axis of the spherical mirror.

We place a point object at [latex]O[/latex]. One incident ray travels along [latex]C[/latex] and remains perpendicular to the surface, thus continuing straight through [latex]PX[/latex]. Another incident ray, [latex]OA[/latex], refracts at point [latex]A[/latex] and bends toward the normal. These two rays intersect at point [latex]I[/latex], forming a virtual image.

Let the angles these rays make with the principal axis be [latex]\alpha[/latex], [latex]\beta[/latex], and [latex]\gamma[/latex] respectively.

By Snell’s Law, the refractive index is:

[latex]\mu = \frac{\sin i}{\sin r}[/latex]

where [latex]i[/latex] is the angle of incidence and [latex]r[/latex] the angle of refraction.

For small angles [latex]i[/latex] and [latex]r[/latex], we approximate:

[latex]\sin i \approx i \quad \text{and} \quad \sin r \approx r[/latex]

so that:

[latex]\mu = \frac{i}{r}[/latex]

leading to:

[latex]i = \mu r[/latex]

Using the Exterior Angle Theoremin [latex]\triangle AOC[/latex]:

[latex]\gamma = i + \alpha \Rightarrow i = \gamma – \alpha[/latex]

Similarly, in [latex]\triangle IAC[/latex]:

[latex]\gamma = \beta + r \Rightarrow r = \gamma – \beta[/latex]

Substitute [latex]i[/latex] and [latex]r[/latex]from these into Snell’s Law:

[latex]\gamma – \alpha = \mu (\gamma – \beta)[/latex]

Since for a spherical surface, [latex]\text{angle} = \frac{\text{arc}}{\text{radius}}[/latex], we can write:

[latex]\alpha = \frac{PA}{OP}, \quad \beta = \frac{PA}{IP}, \quad \gamma = \frac{PA}{CP}[/latex]

Substitute these expressions:

[latex]\frac{PA}{PC} – \frac{PA}{PO} = \mu \left( \frac{PA}{PC} – \frac{PA}{PI} \right)[/latex]

Simplifying by canceling \( PA \) from both sides:

[latex]\frac{1}{PC} – \frac{1}{PO} = \mu \left( \frac{1}{PC} – \frac{1}{PI} \right)[/latex]

Applying the sign convention:

– [latex]PC = -R[/latex] (radius of curvature),

– [latex]PI = -v[/latex] (image distance),

– [latex]PO = -u[/latex] (object distance),

we substitute to get:

[latex]\frac{1}{-R} – \frac{1}{-u} = \mu \left( \frac{1}{-R} – \frac{1}{v} \right)[/latex]

Expanding and rearranging:

[latex]\frac{\mu – 1}{R} = \frac{\mu}{v} – \frac{1}{u}[/latex]

This gives us the desired expression for a concave mirror.

Note on Cartesian Sign Convention:

• Distances measured from the mirror’s pole.
• Positive distances are measured in the direction of incident light, while negative distances are measured opposite to it.
• Heights measured upward from the principal axis are positive, and those downward are negative.