A hemisphere made of a transparent material of variable refractive index has radius 12 cm and centre at origin O. The refractive index of the material of the sphere varies as μ=2424−x, where x is in cm. A ray of light is incident at the point O at an angle θ0 with the x–axis and comes out through a point P on its curved surface. Given θ0≈900, then

Let at any point A of trajectory of ray the tangent to the path of ray makes an angle with x–axis. From Snell’s law 1×sin900=nsinθ ⇒sinθ=1n=24−x24 tanθ=24−x48x−x2 dydx=24−x48x−x2 ∫0ydy=∫0x(24−x)dx48x−x2⇒x2+y2−48x=0 so the path of the ray is circular.At P, x2+y2=R2=144⇒48x=144 ⇒x=3 cm

A hemisphere made of a transparent material of variable refractive index has radius 12 cm and centre at origin O. The refractive index of the material of the sphere varies as μ=2424−x, where x is in cm. A ray of light is incident at the point O at an angle θ0 with the x–axis and comes out through a point P on its curved surface. Given θ0≈900, then

Let at any point A of trajectory of ray the tangent to the path of ray makes an angle with x–axis. From Snell’s law 1×sin900=nsinθ ⇒sinθ=1n=24−x24 tanθ=24−x48x−x2 dydx=24−x48x−x2 ∫0ydy=∫0x(24−x)dx48x−x2⇒x2+y2−48x=0 so the path of the ray is circular.At P, x2+y2=R2=144⇒48x=144 ⇒x=3 cm

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A hemisphere made of a transparent material of variable refractive index has radius 12 cm and centre at origin O. The refractive index of the material of the sphere varies as $μ = \frac{24}{24 - x}$ , where x is in cm. A ray of light is incident at the point O at an angle $θ_{0}$ with the x–axis and comes out through a point P on its curved surface. Given $θ_{0} \approx 90^{0}$ , then

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Path of the ray inside the sphere is circular

Path of the ray inside the sphere is parabolic

x–coordinate of point P is 1.50 cm

x–coordinate of point P is 3 cm

answer is B, C.

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Detailed Solution

Let at any point A of trajectory of ray the tangent to the path of ray makes an angle with x–axis. From Snell’s law

$1 \times \sin 90^{0} = n \sin θ \Rightarrow \sin θ = \frac{1}{n} = \frac{24 - x}{24} \tan θ = \frac{24 - x}{\sqrt{48 x - x^{2}}} \frac{d y}{d x} = \frac{24 - x}{\sqrt{48 x - x^{2}}} \int_{0}^{y} d y = \int_{0}^{x} \frac{(24 - x) d x}{\sqrt{48 x - x^{2}}}$

$\Rightarrow x^{2} + y^{2} - 48 x = 0$ so the path of the ray is circular.
At P, $x^{2} + y^{2} = R^{2} = 144$
$\Rightarrow 48 x = 144 \Rightarrow x = 3 c m$