Question

1. What can one say about the image produced by a thin lens that produces a positive magnification?

a. It is real and inverted

b. It is real and erect.

c.It is virtual and inverted.

d. It is virtual and erect.

2.If the diameter of a lens is reduced, what happens to the magnification produced by the lens?

a. It increases

b. It decreases

c. It is unchanged

Answer 1

Concepts and reason

The concepts used to solve this problem are magnification, image height and the object height.

First, use conditions of the sign convention of magnification produced by thin lens.

Use the expression of magnification in terms of image and object height to prove that there is no relation between the magnification and radius of curvature of lens.

Fundamentals

If the direction of light that falls on lens is from left to right, a coordinate system can be constructed with x and y being the coordinate axes. Then the sign convention is defined as,

Along the x axis, the region to the right of origin is taken as positive and to the left of it is taken as negative.

Along the y axis, the region upwards of origin is taken as positive and the region downwards is taken as negative.

The expression for magnification in terms of image height and object height is,

$m = \frac{{{h_i}}}{{{h_o}}}$

Here, the magnification is $m$ , the image height is ${h_i}$ , and the object height is ${h_o}$ .

The expression for magnification in terms of image distance and object distance is,

$m = - \frac{{{d_i}}}{{{d_o}}}$

Here, the image distance is ${d_i}$ , the object distance is ${d_o}$ .

(1)

The expression for magnification is,

$m = \frac{{{h_i}}}{{{h_o}}}$

The expression for magnification in terms of image distance and object distance is,

$m = - \frac{{{d_i}}}{{{d_o}}}$

The incorrect options are,

a.It is real and inverted

If the magnification produced by a lens is positive, the image cannot be inverted and hence this statement is incorrect.

b.It is real and erect

For a positive magnification, the image distance should be negative and a negative image distance cannot produce real image. Hence this statement is false

c.It is virtual and inverted

The above option is incorrect as the positive magnification cannot produce an inverted image.

The correct option is,

d.It is virtual and erect

The above option is correct because a positive magnification produces an upright image that is virtual.

(2)

The expression for magnification in terms of object and image distance is,

$m = - \frac{{{d_i}}}{{{d_o}}}$

The incorrect options are,

a.It increases

The above option is incorrect because the magnification depends on image height and object height, both of which are independent of radius of curvature.

b.It decreases

The above option is incorrect because the magnification is independent of radius of curvature.

The correct option is

a.It is unchanged

The above option is correct because magnification is independent of the radius of curvature of lens.

Ans: Part 1

The image produced by the thin lens is virtual and erect.