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 scree 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 $(2 f, 2 f)$
For the graph to intersect $y = x$ line,
the value of $| v |$ and $| u |$ must be equal.

$u = - u$ ; $v = u$

From lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{u} + \frac{1}{u} = \frac{1}{f}$

$u = - 2 f$ and $v = + 2 f$

The coordinates are $(2 f, 2 f)$

option $A$ is correct.

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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)For the graph to intersect y=x line,the value of |v| and |u| must be equal.u=−u ; v=uFrom lens formula, 1v−1u=1f1u+1u=1fu=−2f and v=+2fThe coordinates are (2f,2f)option A is correct.