A hemisphere made of a transparent material of variable refractive index has radius 12 cm and center 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 which of the following statement(s) is (are) true

Let at any point A of trajectory of ray the tangent to the path of ray makes an angle θ with x−axis. From snell’s law1×sin900=nsinθ ⇒sinθ=1n=24−x24tanθ=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=3cm

A hemisphere made of a transparent material of variable refractive index has radius 12 cm and center 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 which of the following statement(s) is (are) true

Let at any point A of trajectory of ray the tangent to the path of ray makes an angle θ with x−axis. From snell’s law1×sin900=nsinθ ⇒sinθ=1n=24−x24tanθ=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=3cm

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A hemisphere made of a transparent material of variable refractive index has radius 12 cm and center 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 $x - a x i s$ and comes out through a point P on its curved surface. Given $θ_{0} \approx 90^{0}$ , then which of the following statement(s) is (are) true

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x – coordinate of point P is $3 c m$

Path of the ray inside the sphere is circular

Path of the ray inside the sphere is parabolic.

y – coordinate of point P is approximately $11.6 c m$

answer is B, C, D.

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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 $θ$ with $x - a x i s$ .

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, $x^{2} + y^{2} = R^{2} = 144$
$\Rightarrow 48 x = 144 \Rightarrow x = 3 c m$