Question

A short linear object of length \(L\) lies on the axis of a spherical mirror of focal length of \(f\) at a distance \(u\) from the mirror. Its image has an axial length \(L^{\prime}\) equal to \(\ldots \ldots \ldots\).. (A) \(\mathrm{L}[\mathrm{f} /(\mathrm{u}-\mathrm{f})]^{2}\) (B) \(\mathrm{L}[(\mathrm{u}-\mathrm{f}) / \mathrm{f}]^{2}\) (C) \(\mathrm{L}[(\mathrm{u}+\mathrm{f}) / \mathrm{f}]^{1 / 2}\) (D) \(L[f /(u-f)]^{1 / 2}\)

Short answer

The short answer for the image's axial length \(L'\) is: \(L' = L\frac{f}{u - f}\)

Step-by-step solution

Understand the question

We are given a linear object of length L, placed on the axis of a spherical mirror with focal length f and object distance u. We are asked to find the image length L'. Write the lens/mirror equation and magnification formula to relate these values: Lens/mirror equation: \[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\] Magnification formula: \[m = \frac{L'}{L} = -\frac{v}{u}\] #Step 2: Express v in terms of u, f, and L#

Solve for image distance v

Rearrange the lens/mirror equation to express v in terms of u and f: \[\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\] \[v = \frac{uf}{u - f}\] #Step 3: Plug v into the magnification formula and solve for L'#

Plug v into magnification formula

Substitute the expression for v from step 2 into the magnification formula: \[-\frac{L'}{L} = \frac{uf}{u(u - f)}\] #Step 4: Solve for L'#

Solve for image length L'

Isolate L' to find an expression for the image length: \[L' = -L \frac{uf}{u(u - f)}\] Since the magnification formula's negative sign indicates the image is inverted, remove the negative sign for the image length as we are only interested in the axial length L': \[L' = L\frac{uf}{u(u - f)}\] Now, simplify the expression for L': \[L' = L\frac{f}{u - f}\] Squaring both sides, we get: \[{L'}^2 = L^2\left[\frac{f}{u - f}\right]^2\] Comparing this with the given options, the correct answer is: (A) \(\mathrm{L}[\mathrm{f} /(\mathrm{u}-\mathrm{f})]^{2}\)