
A transparent (glass) sphere has a small, opaque dot at its centre. Does the apparent position of a dot appear to be the same as its actual position when observed from outside?
Answer
477.3k+ views
Hint: According to Snell’s law the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. All the rays that come out from an object that is kept at the centre of a transparent sphere will be normal to the surface of the sphere. The light will be reflected inside the sphere.
Complete step by step answer:
When light falls normally on the boundary surface then the angle if incidence will be $0$,
We will apply the lens formula considering the refractive indices of the medium involved
According to this formula;
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{\left( {{\mu _2} - {\mu _1}} \right)}}{R}$ ……………………………………… (i)
As we know that refractive index of glass is will be constant ${\mu _1} = \mu $,
${\mu _2}$(refractive index of surrounding) will be 1, $u = - R$ = {Radius of sphere} and \[R = - R\] is the radius of sphere
Substituting these values in equation (i)
We get,
$\dfrac{1}{v} - \dfrac{\mu }{{ - R}} = \dfrac{{\left( {1 - \mu } \right)}}{{ - R}}$………………………………………………(ii)
On solving above equation, we get,
$\dfrac{1}{v} = \dfrac{1}{{ - R}}$ …………………………………………………………….(iii)
From above relation, we can say that the "there will be no change in the position of an opaque dot at the centre of a transparent glass sphere when a person observes it from the outside."
Note:
When light falls normally on the boundary surface, then we have angle of incidence $i = 0$ and
According to Snell's law \[{\mu _1}\sin i\; = {\mu _2}\sin r.\] So, there will be no refraction, i.e. r is also zero. Thus, the light will be totally internally reflected.
Hence the object will be observed at the same position from outside, and this and will be independent of the refractive index of the sphere
Complete step by step answer:
When light falls normally on the boundary surface then the angle if incidence will be $0$,
We will apply the lens formula considering the refractive indices of the medium involved
According to this formula;
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{\left( {{\mu _2} - {\mu _1}} \right)}}{R}$ ……………………………………… (i)
As we know that refractive index of glass is will be constant ${\mu _1} = \mu $,
${\mu _2}$(refractive index of surrounding) will be 1, $u = - R$ = {Radius of sphere} and \[R = - R\] is the radius of sphere
Substituting these values in equation (i)
We get,
$\dfrac{1}{v} - \dfrac{\mu }{{ - R}} = \dfrac{{\left( {1 - \mu } \right)}}{{ - R}}$………………………………………………(ii)
On solving above equation, we get,
$\dfrac{1}{v} = \dfrac{1}{{ - R}}$ …………………………………………………………….(iii)
From above relation, we can say that the "there will be no change in the position of an opaque dot at the centre of a transparent glass sphere when a person observes it from the outside."
Note:
When light falls normally on the boundary surface, then we have angle of incidence $i = 0$ and
According to Snell's law \[{\mu _1}\sin i\; = {\mu _2}\sin r.\] So, there will be no refraction, i.e. r is also zero. Thus, the light will be totally internally reflected.
Hence the object will be observed at the same position from outside, and this and will be independent of the refractive index of the sphere
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Types of lever in which effort is in between fulcrum class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

A two input XOR Gate produces a high output only when class 12 physics CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE
