Question

If the object begins to move the speed \[{V_{\text{o}}}\], what will be the speed of its image formed by a spherical mirror with focal length \[f\] ?

Answer 1

Hint: Use the mirror formula and then differentiate the whole expression with respect to the small change in object distance. Find the ratio of the image and object distance. Obtain this ratio by multiplying object distance on both sides of the formula. To find the speed of the image, you must differentiate with respect to time.

Complete step by step answer:
The speed of the object is given as \[{V_{\text{o}}}\] .
Mirror formula is an equation that compares the distance of an object and is known as a mirror equation for image distance with focal length. The mirror formula is given by,
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\] …… (i)
We know that the focal length is constant for a given mirror.

Differentiate equation (i) with respect to \[du\], it gives:
$- \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{du}} - \dfrac{1}{{{u^2}}}\dfrac{{du}}{{du}} = 0 \\
- \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{du}} = \dfrac{1}{{{u^2}}} \\
dv = - {\left( {\dfrac{v}{u}} \right)^2}du \\ $ …… (ii)
Multiply equation (i) with \[u\]
$\dfrac{u}{v} + 1 = \dfrac{u}{f} \\
\implies \dfrac{u}{v} = \dfrac{u}{f} - 1 \\
\implies \dfrac{u}{v} = \dfrac{{u - f}}{f} \\
\implies \dfrac{v}{u} = \dfrac{f}{{u - f}} \\$ …… (iii)
Now place the value of equation (iii) in equation (ii)
Therefore,
\[dv = - {\left( {\dfrac{f}{{u - f}}} \right)^2}du\]
Divide both side of the above equation with \[dt\], we get
\[\dfrac{{dv}}{{dt}} = - {\left( {\dfrac{f}{{u - f}}} \right)^2}\dfrac{{du}}{{dt}}\] …… (iv)
According to the question,
\[\dfrac{{du}}{{dt}} = {V_{\text{o}}}\]
Let us consider the speed of image as:
\[\dfrac{{dv}}{{dt}} = {V_{\text{i}}}\]
Therefore, equation (iv) becomes,
\[{V_{\text{i}}} = - {\left( {\dfrac{f}{{u - f}}} \right)^2}{V_{\text{o}}}\]

Additional Information:
A mirror is a reflecting surface that does not allow light to pass in, and instead bounces it off, forming an image. Flat and so-called plane mirrors are the most common mirrors. These mirrors are created by placing a thin film of silver nitrate or aluminium behind a flat sheet of glass. A mirror with a texture that is convex or concave and forms a part of the real sphere.
A convex structure is the reverse of a concave shape. In the case of convex mirrors, the inner layer is polished, with the outer layer acting as a reflecting surface. However, in the case of a concave mirror, the outer layer is polished, with the inner layer acting as a reflecting surface.

Note:
This problem can be solved if you understand what calculus (differentiation) is. First of all, you must obtain the changes in image and object distance. While doing so, remember that differentiation of constant is always equal to zero. After doing the above step correctly, then you go for substituting the required values. Since, the question mentions speed, don’t forget to differentiate with respect to time.