In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $u$ and the image distance $v$ , from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of $45^{o}$ with the $x$ -axis meets the experimental curve at P. The coordinates of $P$ will be

(2 f, 2 f)

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(\frac{f}{2}, \frac{f}{2})

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(f, f)

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(4 f, 4 f)

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Solution

The correct option is A $(2 f, 2 f)$
The mentioned straight line is the line $x = y$
On this line $u = v$
Following sign convention,
$u = - u, v = v = u$
Substituting this into the lens equation, we get
$\frac{1}{u} - \frac{- 1}{u} = \frac{1}{f}$
Solving this, we get $u = v = 2 f$
Therefore the point is $(2 f, 2 f)$

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Similar questions

Q. Draw a diagram to represent the nature, position and relative size of the image formed by a i) convex lens for the object placed at

2 F

ii) concave lens for the object placed between

F

and

2 F

A virtual, erect and magnified image of an object is to be obtained with a convex lens. For this purpose, the object should be placed : (a) between 2F and infinity (b) between F and optical centre (c) between F and 2F (d) at F

An object is placed at a distance of $\frac{f}{2}$ from a convex lens. The image will be formed:

Q. (a) Draw ray diagrams to show the formation of a three times magnified (i) real image (ii) virtual image, of an object kept in front of a converging lens. Mark the positions of object, F, 2F, O and position of image clearly in the diagram.
(b) An object of size 5 cm is kept at a distance of 25 cm from the optical centre of a converging lens of focal length 10 cm. Calculate the distance of the image from the lens and size of the image.

Q. An object placed between

F

and

2 F

of a convex lens will produce an image

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The correct option is A (2f,2f)The mentioned straight line is the line x=yOn this line u=vFollowing sign convention,u=−u,v=v=uSubstituting this into the lens equation, we get1u−−1u=1fSolving this, we get u=v=2fTherefore the point is (2f,2f)