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Chapter 35: Problem 2880

A convex lens of focal length \(\mathrm{f}\) is placed somewhere in between an object and a screen. The distance between the object and the screen is \(\mathrm{x}\). If the numerical value of the magnification product by the lens is \(\mathrm{m}\), What is the focal length of the lens? (A) \(\left[\mathrm{mx} /(\mathrm{m}-1)^{2}\right]\) (B) \(\left[\mathrm{mx} /(\mathrm{m}+1)^{2}\right]\) (C) \(\left[(m-1)^{2} / \mathrm{m}\right] \mathrm{x}\) (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\)

Short Answer

Expert verified
The short answer is: The focal length of the lens is given by (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\).

Step by step solution

01

Write down the lens formula

The lens formula is given by \(\frac{1}{\mathrm{f}} = \frac{1}{\mathrm{v}} - \frac{1}{\mathrm{u}}\), where f is the focal length of the lens and v, u are the image and object distances from the lens respectively.
02

Write down the magnification formula

The magnification formula for lenses is given by \(\mathrm{m} = - \frac{\mathrm{v}}{\mathrm{u}}\). This indicates that the ratio of the image distance to the object distance will be equal to the magnification of the lens.
03

Substitute the distance between object and screen into the formula

It is given that the distance between object and screen (x) is equal to the sum of the object and image distances i.e. x = u + v.
04

Express v in terms of u and m from the magnification formula

Using the magnification formula, we have v = -mu. Substitute this in x = u + v. So, x = u - mu.
05

Calculate the value of u in terms of m and x

Using the above equation, we can calculate u as follows: u = x / (m - 1).
06

Calculate the value of v in terms of m and x

Now substitute the value of u from the previous step into the magnification formula v = -mu to find the value of v. v = -mx/ (m - 1).
07

Substitute the values of u and v in the lens formula

Now substitute these values of u and v in terms of m and x into the lens formula: \(\frac{1}{\mathrm{f}} = \frac{1}{- \mathrm{mx} / (\mathrm{m} - 1)} - \frac{1}{\mathrm{x} / (\mathrm{m} - 1)}\).
08

Solve for f

Now, solve for f: f = \(\frac{(\mathrm{m} - 1)^2}{\mathrm{m}} \cdot \mathrm{x}\). Hence, the correct option is (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\).

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Most popular questions from this chapter

A convex lens of focal length \(f\) produces a real image \(x\) times the size of an object, Then what is the distance of the object from the lens? (A) \((\mathrm{x}+1) \mathrm{f}\) (B) \((\mathrm{x}-1) \mathrm{f}\) (C) \([(\mathrm{x}+1) / \mathrm{x}] \mathrm{f}\) (D) \([(x-1) / x] f\)

An object is placed at a distance of \((\mathrm{f} / 2)\) the from a convex lens the image will be.... (A) at \(\mathrm{f}\), real and inverted (B) at,\((3 \mathrm{f} / 2)\) real and inverted (C) at one of the foci, virtual and double its size (D) at \(2 \mathrm{f}\), virtual and erect.

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) \(L[f /(u-f)]^{1 / 2}\)

A concave lens of focal length \(\mathrm{f}\) forms an image which is n times the size of the object. What is the distance of the object from the lens? (A) \((1+\mathrm{n}) \mathrm{f}\) (B) \((1-\mathrm{n}) \mathrm{f}\) (C) \([(1-\mathrm{n}) / \mathrm{n}] \mathrm{f}\) (D) \([(1+\mathrm{n}) / \mathrm{n}] \mathrm{f}\)

A concave mirror of focal length \(\mathrm{f}\) produces an images n times the size of the object. If the image is real then What is the distance of the object from the mirror? (A) \((\mathrm{n}+1) \mathrm{f}\) (B) \([(\mathrm{n}-1) / \mathrm{n}] \mathrm{f}\) (C) \((\mathrm{n}-1) \mathrm{f}\) (D) \([(\mathrm{n}+1) / \mathrm{n}] \mathrm{f}\)
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