
A transparent sphere of radius R has a cavity of radius $\dfrac{R}{2}$ as shown in Fig. The refractive index of the sphere if a parallel beam of light falling on the left surface focuses at point P is $\dfrac{3+\sqrt{5}}{x}$. Find out the value of X?
Answer
162.3k+ views
Hint:We have to find the refractive index of the sphere if a parallel beam of light falling on the left surface focuses on a point. It is given that this transparent sphere of radius R has a cavity of radius $\dfrac{R}{2}$ . Using the equation connecting image distance, refractive index and radius we can find the value of x.
Formula used:
Let refractive index be μ and v be image distance and if the light is falling parallel then we have the equation as:
$\dfrac{\mu }{v}-\dfrac{1}{\infty }=\dfrac{\mu }{\mu -1}$
Complete step by step solution:
We have a transparent sphere of radius R and it has a cavity with radius $\dfrac{R}{2}$. Refractive index of the sphere is given as $\dfrac{3+\sqrt{5}}{x}$. This refractive index is measured when a parallel beam of light falling on the left surface focuses at point P. All conditions are given and we have to find the value of x. That is, we have to find a refractive index. When a parallel beam of light is falling on the left surface, then object distance will be infinity.
Let refractive index be μ and v be image distance and if the light is falling parallel then we have the equation as:
$\dfrac{\mu }{v}-\dfrac{1}{\infty }=\dfrac{\mu }{\mu -1}$
That is,
$v=\dfrac{\mu R}{\mu -1}$
For finding object distance at second boundary, we have:
$u=\dfrac{\mu R}{\mu -1}-R=\dfrac{R}{\mu -1}$
Hence the equation for refraction at second boundary is
$\dfrac{\mu }{-u}+\dfrac{1}{R}=\dfrac{1-\mu }{\dfrac{R}{2}}$
On substituting the values, we get:
$\dfrac{1}{R}=\dfrac{2(1-\mu )}{R}+\dfrac{\mu }{\dfrac{R}{\mu -1}}$
On further solving we get a quadratic equation in terms of refractive index (μ) as:
${{\mu }^{2}}-3\mu +1=0$
On solving this quadratic equation, we get refractive index as:
Refractive index, $\mu =\dfrac{3+\sqrt{5}}{2}$
Therefore, the answer is $x=2$
Notes: Here on solving quadratic equations we get two values for the refractive index. But the refractive index is already given, we only have to find the denominator so we ignore another value. In this question we consider refraction at two surfaces.
Formula used:
Let refractive index be μ and v be image distance and if the light is falling parallel then we have the equation as:
$\dfrac{\mu }{v}-\dfrac{1}{\infty }=\dfrac{\mu }{\mu -1}$
Complete step by step solution:
We have a transparent sphere of radius R and it has a cavity with radius $\dfrac{R}{2}$. Refractive index of the sphere is given as $\dfrac{3+\sqrt{5}}{x}$. This refractive index is measured when a parallel beam of light falling on the left surface focuses at point P. All conditions are given and we have to find the value of x. That is, we have to find a refractive index. When a parallel beam of light is falling on the left surface, then object distance will be infinity.
Let refractive index be μ and v be image distance and if the light is falling parallel then we have the equation as:
$\dfrac{\mu }{v}-\dfrac{1}{\infty }=\dfrac{\mu }{\mu -1}$
That is,
$v=\dfrac{\mu R}{\mu -1}$
For finding object distance at second boundary, we have:
$u=\dfrac{\mu R}{\mu -1}-R=\dfrac{R}{\mu -1}$
Hence the equation for refraction at second boundary is
$\dfrac{\mu }{-u}+\dfrac{1}{R}=\dfrac{1-\mu }{\dfrac{R}{2}}$
On substituting the values, we get:
$\dfrac{1}{R}=\dfrac{2(1-\mu )}{R}+\dfrac{\mu }{\dfrac{R}{\mu -1}}$
On further solving we get a quadratic equation in terms of refractive index (μ) as:
${{\mu }^{2}}-3\mu +1=0$
On solving this quadratic equation, we get refractive index as:
Refractive index, $\mu =\dfrac{3+\sqrt{5}}{2}$
Therefore, the answer is $x=2$
Notes: Here on solving quadratic equations we get two values for the refractive index. But the refractive index is already given, we only have to find the denominator so we ignore another value. In this question we consider refraction at two surfaces.
Recently Updated Pages
How To Find Mean Deviation For Ungrouped Data

Difference Between Molecule and Compound: JEE Main 2024

Ammonium Hydroxide Formula - Chemical, Molecular Formula and Uses

Difference Between Area and Surface Area: JEE Main 2024

Difference Between Work and Power: JEE Main 2024

Difference Between Acetic Acid and Glacial Acetic Acid: JEE Main 2024

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
