Question
A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is _________ cm.
Refraction at Convex Spherical Surface
This problem requires calculating the image position formed by a convex spherical surface when parallel light rays are incident on it.
Given Data
- Medium 1 (Air): Refractive index, $n_1 = 1.0$
- Medium 2 (Glass): Refractive index, $n_2 = 1.5$
- Surface type: Convex spherical
- Radius of curvature: $R = +50$ cm (Positive because the centre of curvature is on the side of the refracted light for a convex surface)
- Incident rays: Parallel beam travelling in air, so the object distance is at infinity, $u = +\infty$
Refraction Formula Application
Use the spherical refraction formula:
$ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $
Substitute the known values:
$ \frac{1.5}{v} - \frac{1.0}{+\infty} = \frac{1.5 - 1.0}{+50 \text{ cm}} $
Since $\frac{1}{\infty} \to 0$, the equation simplifies:
$ \frac{1.5}{v} = \frac{0.5}{50 \text{ cm}} $
$ \frac{1.5}{v} = \frac{1}{100 \text{ cm}} $
Solve for $v$, the image distance from the surface:
$ v = 1.5 \times 100 \text{ cm} = 150 \text{ cm} $
Distance from Centre of Curvature
The image distance $v = 150$ cm is measured from the spherical surface (pole). The question asks for the distance $x$ from the centre of curvature.
The centre of curvature is at a distance $R$ from the surface.
Therefore, the distance $x$ from the centre of curvature is:
$ x = |v - R| $
Substitute the values:
$ x = |150 \text{ cm} - 50 \text{ cm}| $
$ x = 100 \text{ cm} $
Final Result
The parallel rays converge at a distance of $100$ cm from the centre of curvature.
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