Answer

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**Hint:**Since the light rays are parallel to each other, then object distance can be regarded as infinity. The point of focus of the ray is the image distance.

Formula used: In this solution we will be using the following formulae;

\[\dfrac{n}{v} - \dfrac{1}{u} = \dfrac{{n - 1}}{R}\]where \[n\] is the refractive index of the glass, \[v\] is the image distance, and \[u\] is the object distance in the first case, \[R\] is the radius of the glass, this is a thick lens equation for light striking the curve surface the Plano-convex.

\[\dfrac{1}{v} - \dfrac{n}{u} = \dfrac{{1 - n}}{R}\] this is the thick lens equation for when incident light strikes the flat surface of a Plano-convex lens.

**Complete Step-by-Step solution:**

For the first case, the light is incident on the curve side, hence, the governing equation can be given be given by

\[\dfrac{n}{{{v_1}}} - \dfrac{1}{{{u_1}}} = \dfrac{{n - 1}}{R}\] where \[n\] is the refractive index of the glass, \[{v_1}\] is the image distance in the first case, and \[{u_1}\] is the object distance in the first case, \[R\] is the radius of the glass.

Since the light rays were parallel to each other, then the object distance can be regarded as infinity, hence

\[\dfrac{n}{{{v_1}}} - \dfrac{1}{\infty } = \dfrac{{n - 1}}{R}\]

\[ \Rightarrow \dfrac{n}{{{v_1}}} = \dfrac{{n - 1}}{R}\]

The point of convergence is the image distance, hence as given in question

\[\dfrac{n}{{\dfrac{{2R}}{m}}} = \dfrac{{n - 1}}{R}\]

Hence, by simplifying and cancelling \[R\] we have

\[mn = 2\left( {n - 1} \right)\]

For the second case, the light is incident on the flat part of the glass, hence, the governing equation is given by

\[\dfrac{1}{{{v_2}}} - \dfrac{n}{{{u_2}}} = \dfrac{{1 - n}}{R}\]

Hence, since light is parallel again, and the point of convergence is \[\dfrac{R}{{m - 1}}\], hence, by substitution, we have

\[\dfrac{1}{{\dfrac{R}{{m - 1}}}} - \dfrac{n}{\infty } = \dfrac{{1 - n}}{R}\]

By simplifying, we get

\[\dfrac{{m - 1}}{R} = \dfrac{{1 - n}}{{ - R}}\]

\[ \Rightarrow m - 1 = 1 - n\]

Then, we see that

\[m = 2 - n\]

Inserting this expression into the equation \[mn = 2\left( {n - 1} \right)\], we get

\[\left( {2 - n} \right)n = 2\left( {n - 1} \right)\]

\[ \Rightarrow 2n - {n^2} = 2n - 2\]

By cancelling and simplifying, we get

\[{n^2} = 2\]

\[ \Rightarrow n = 1.4\]

**Note:**For clarity, observe that the equation \[\dfrac{n}{v} - \dfrac{1}{u} = \dfrac{{n - 1}}{R}\], for example, is very similar to the lens maker equation \[\dfrac{1}{f} = n - 1\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)\]. But since the surface is flat, we have that one of the radius be equal to infinity reducing the equation to \[\dfrac{1}{f} = n - 1\left( {\dfrac{1}{{{R_1}}}} \right)\] then \[\dfrac{1}{f}\] can be proven to be \[\dfrac{n}{v} - \dfrac{1}{u}\]. Similar case for the second equation.

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