Answer

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**Hint:**Use the mirror’s formula given as $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ where u is the object’s distance, v is the image’s distance and f is the focal length of the mirror. To calculate the value of f, use the relation $f=\dfrac{R}{2}$ where R is the radius of the concave mirror. Find the value of $\dfrac{-v}{u}$ using the data given and equate it with $\dfrac{h'}{h}$ where h is the height of the object and h’ is the height of the image. Consider the diameter as the object and calculate the value of h’ to get the answer. Take the value of u negative in the mirror’s formula.

**Complete step by step solution:**

Here we have been provided with the diameter of the moon whose image is formed by a concave mirror. We are asked to find the diameter of the image of the moon.

Now, we know that the mirror’s formula is given as $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ where u is the object’s distance, v is the image’s distance and f is the focal length of the mirror. Here f is given as $f=\dfrac{R}{2}$ where R is the radius of curvature of the concave mirror. Therefore, substituting in the formulas we get,

$\Rightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{2}{R}$

Multiplying both the sides with u we get,

$\begin{align}

& \Rightarrow 1+\dfrac{u}{v}=\dfrac{2u}{R} \\

& \Rightarrow \dfrac{u}{v}=\dfrac{2u}{R}-1 \\

\end{align}$

Substituting the values of u and f after converting them into the unit meter and using the proper sign convention we get,

$\begin{align}

& \Rightarrow \dfrac{u}{v}=-\left( \dfrac{2\times 3.5\times {{10}^{5}}\times {{10}^{3}}}{3} \right)-1 \\

& \Rightarrow \dfrac{u}{v}=-\left( \dfrac{7\times {{10}^{8}}}{3} \right)-1 \\

& \Rightarrow \dfrac{-u}{v}=\left( \dfrac{7\times {{10}^{8}}}{3} \right)+1 \\

\end{align}$

Here we can neglect 1 because $\left( \dfrac{7\times {{10}^{8}}}{3} \right)$ is a very large quantity, so we get,

$\begin{align}

& \Rightarrow \dfrac{-u}{v}=\left( \dfrac{7\times {{10}^{8}}}{3} \right) \\

& \Rightarrow \dfrac{-v}{u}=\dfrac{3\times {{10}^{-8}}}{7} \\

\end{align}$

Now, we know that the magnification by a plane mirror is given as $\dfrac{-v}{u}$ and in terms of the height of the image (h’) and height of the object (h) as $\dfrac{h'}{h}$, so considering the diameter of the moon as the object we have,

$\Rightarrow \dfrac{-v}{u}=\dfrac{h'}{h}$

Substituting the given values we get,

$\begin{align}

& \Rightarrow \dfrac{h'}{3500\times {{10}^{3}}}=\dfrac{3\times {{10}^{-8}}}{7} \\

& \Rightarrow h'=\dfrac{3\times {{10}^{-8}}}{7}\times 3500\times {{10}^{3}} \\

& \Rightarrow h'=15\times {{10}^{-3}}m \\

\end{align}$

We know that 1 mm = 0.001 m so we get,

$\Rightarrow h'=15mm$

Therefore, the diameter of the image of the moon is 15 mm.

**So, the correct answer is “Option c”.**

**Note:**You can say that the image of the moon will be formed at the focus of the mirror because the moon can be assumed to be situated at infinity. So there can be another method by which we can solve the question. In that method properties of the similar triangles will be used, however the above process that we have applied is nothing but the physical interpretation of the properties of similar triangles. Remember the sign convention used in mirror and lens formula.

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