Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

When the distance between the object and the screen is more than 4f, we can obtain the image of the object on the screen for the two positions of a convex lens of focal length f. It is called displacement method. In one case the image is magnified. If $I_1$ and $I_2$ be the sizes of the two images, then the size of the object is:

Options:

$(I_1 + I_2)/2$

$I_1 − I_2$

$\sqrt{(I_1 I_2)}$

$\sqrt{(I_1/I_2)}$

Correct Answer:

$\sqrt{(I_1/I_2)}$

Explanation:

If $I_1 = mI_0$ where m is magnification and $I_0$ is actual size

$I_2 = I_0/m$

$∴ I_0 = \sqrt{I_1 I_2}$