
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
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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.
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