Question 270: Ray Optics

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 $x \mathrm{~cm}$ from the centre of the curvature of the spherical surface. The value of $x$ is $\_\_\_\_$ cm .

Accepted Answer

Imagine a point object $O$ placed on the principal axis of a convex spherical surface that separates two media of refractive indices $\mu_1$ and $\mu_2$.
Here the pole of the surface be P and its centre of curvature be C.
Using the formula for refraction at a spherical surface :
$$ \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} $$
The incident beam is parallel, which means the light is coming from an object at infinity. Therefore, the object distance $u=-\infty$.
The refractive index of the first medium (air) is $\mu_1=1.0$.
The refractive index of the second medium (glass) is $\mu_2=1.5$.
The surface is convex facing the air. This means its centre of curvature lies inside the glass, in the direction of the traveling light. Thus, the radius of curvature is positive: $\mathrm{R}=+50 \mathrm{~cm}$.
Let's plug these into our derived equation to find the image position (v) :
$$ \frac{1.5}{v}-\frac{1.0}{-\infty}=\frac{1.5-1.0}{+50} $$
$$\Rightarrow $$ $\frac{1.5}{v}-0=\frac{0.5}{50}$
$$\Rightarrow $$ $\frac{1.5}{v}=\frac{1}{100}$
$$\Rightarrow $$ $$ \mathrm{v}=150 \mathrm{~cm} $$
So, the parallel rays will converge to form an image at a distance of 150 cm to the right of the pole of the spherical surface.
Pole $(P)$ is at 0 cm . and the centre of curvature $(C)$ is at +50 cm .
Image point / Convergence point (I) is at +150 cm .
So, the distance between C and I is
$$ x=v-R $$
$$\Rightarrow $$ $$ \mathrm{x}=150 \mathrm{~cm}-50 \mathrm{~cm}=100 \mathrm{~cm} $$
Therefore, the distance from the centre of curvature is 100 cm .
Hence, the correct answer is $\mathbf{1 0 0}$.

Ray Optics

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