# A parallel beam of light enters a clear plastic bead 2.5 cm in diameter and index 1.44. At what point beyond the bead are these rays brought to a focus?

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Hint: Here in this question the beam of light that enters the bead is parallel meaning the light rays are coming from at a distance of infinity. So, the object is at infinity. Since we know the object distance and the radius of the bead, we can apply this formula to calculate the image distance.
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Where $\mu$ is the refractive index of the medium, $R$ is the radius of the bead and are $v$ and $u$image and object distances.

The object is placed at a distance behind the bead as per the convention that the light rays move from negative to positive direction. The object distance is considered to be negative.
So, our object distance is minus infinity.
$u = - \infty$
The refractive index of air is 1 and the bead is 1.44
${\mu _1} = 1$
${\mu _2} = 1.44$
Since the diameter of the bead is 2.5 cm the radius of the bead becomes 1.25 cm
$R = 1.25cm$
Now we substitute the value of the above variables into the equation
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
$\dfrac{{1.44}}{v} - \dfrac{1}{{ - \infty }} = \dfrac{{1.44 - 1}}{{1.25}}$
As $\dfrac{1}{\infty }$ is equal to zero the above equation becomes
$\dfrac{{1.44}}{v} - 0 = \dfrac{{1.44 - 1}}{{1.25}}$
$\dfrac{{1.44}}{v} = \dfrac{{0.44}}{{1.25}}$
Now by rearranging the terms of the equation we get the value of the image distance.
$v = \dfrac{{1.44 \times 1.25}}{{0.44}} = 4.09cm$
So, the distance of the image formed is 4.09cm from the centre of the bead. And all the rays are focused on this particular point.

Note:
Since we are following the sign convention that light moves from the negative side to the positive side, we must stick with this for solving the entire equation. Placing a wrong sign in place may lead to serious miscalculations and we will eventually land up with the wrong answer. Proper sign convention should be followed when we are solving the questions related to lens and mirror.