If the distance from a converging lens to the object is less than the focal length of the lens, the image will be
A. real, inverted, and larger than the object.
B. real, inverted, and smaller than the object.
C. virtual, upright, and larger than the object.
D. virtual, upright, and smaller than the object.

The correct answer is letter C. virtual, upright, and larger than the object. If the distance from a converging lens to the object is less than the focal length of the lens, the image will be virtual, upright, and larger than the object.

Answer Link

aristocles aristocles · Answer 2 · 2019-05-13T03:02:05+02:00

Answer:

C. virtual, upright, and larger than the object.

Explanation:

As we know by lens formula

[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]

by solving above the position of image is given as

[tex]d_i = \frac{(d_0)(f)}{d_0 - f}[/tex]

now the magnification is given as

[tex]m = \frac{d_i}{d_0}[/tex]

[tex]m = \frac{f}{d_0 - f}[/tex]

now here given that distance of object from lens is less than its focal length

so here we can say

[tex]m = -\frac{f}{f - d_0}[/tex]

so it shows magnification is greater than 1 in magnitude and also its negative so it must be virtual image and magnified image with upright position