Question

The radius of a glass ball is$5cm$ . There is an air bubble at $1cm$ from the center of the ball and the refractive index of glass is$1.5$ . The position of the image viewed from the surface near the bubble is.

Answer 1

In this question first of all we should know about the mean of the radius of any component and also about the refractive index for the glass. Suppose an imaginary ball that has a certain radius in an air bubble far from the center of the ball. Use the formula of the radius of curvature in terms of the object-distance and image-distance and put the given values correctly. Note that, the object distance should be the difference between the two positions given in the problem.

Formula used:
\left\\{ {\left( {\dfrac{{{\mu _2}}}{v}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{{\mu _1}}}{u}} \right)} \right\\} = {\text{ }}\left( {\dfrac{{{\mu _2} - {\text{ }}{\mu _1}}}{R}} \right)
${\mu _1} =$ the refractive index of the air
${\mu _2} =$ the refractive index of the glass
$v =$ image distance.
$u =$ object distance = the difference between the radius of the ball and the position of the bubble.
The radius of the ball $R$

Complete answer:
First, we have to find the object's distance.
Given that,
The radius of the glass ball $R = 5cm$
Distance of air bubble from the center of the ball $= 1cm$
So, the object distance should be, $u = 5 - 1 = 4cm$
We know,
\left\\{ {\left( {\dfrac{{{\mu _2}}}{v}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{{\mu _1}}}{u}} \right)} \right\\} = {\text{ }}\left( {\dfrac{{{\mu _2} - {\text{ }}{\mu _1}}}{R}} \right)
The refractive index of the air medium${\mu _1} = 1$
The refractive index of the glass medium ${\mu _2} = 1.5$
We have to find the image distance $v$.
Putting the values in the above formula
\left\\{ {\left( {\dfrac{1}{v}} \right){\text{ }} - {\text{ }}\left( {\dfrac{{1.5}}{{ - 4}}} \right)} \right\\} = {\text{ }}\left( {\dfrac{{1 - {\text{ 1}}{\text{.5}}}}{R}} \right)[negative sign is for the direction]
$\Rightarrow \dfrac{1}{v} + \dfrac{{1.5}}{4} = {\text{ }}\dfrac{{0.5}}{5}$
$\Rightarrow \dfrac{1}{v} = {\text{ }}\dfrac{1}{{10}} - \dfrac{{1.5}}{4}$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{{15}}{{40}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{{ - 11}}{{40}}$
$\Rightarrow v = - \dfrac{{40}}{{11}} = - 3.63$
The position of the image viewed from the surface near the bubble is $3.63cm$

Note:
Refractive index: - Refractive index, also called the index of refraction, the measure of the bending of a ray of light when passing from one medium into another. If $i$ is the angle of incidence of a ray in the vacuum and $r$ is the angle of refraction, the refractive index $\mu$ is defined as the ratio of the sine of the angle of incidence to the sine of the angle of refraction; i.e$\mu = \dfrac{{sin\;i\;}}{{sin\;r}}$ . Refractive index is also equal to the velocity of light of a given wavelength in empty space divided by its velocity v in a substance, or $\mu = \dfrac{c}{v}$