Question

(iii) What is the formula for magnification obtained with a lens?
(a) Ratio of height of image to height of object
(b) Double the focal length.
(c) Inverse of the radius of curvature.
(d) Inverse of the object distance.​

Answer 1

Step-by-step explanation:

Students will learn how light behaves when passing through converging and diverging lenses. Students will also learn how to do ray tracing diagrams and calculate image distances and magnification using the lens' maker's equation.

Key Equations

1f=1v−1u

The len's makers's equation

Where f is the focal length of the lens, u is the distance of the object from the lens and v is the distance the image is formed from the lens.

M=vu

The size of an object’s image is larger (or smaller) than the object itself by its magnification, M. The level of magnification is proportional to the ratio of v and u. An image that is double the size of the object would have magnification M=2.

Example 1

You have a converging lens of focal length 2 units. If you place an object 5 units away from the lens, (a) draw a ray diagram of the situation to estimate where the image will be and (b) list the charactatistics of the image. Finally (c) calculate the position of the image. A diagram of the situation is shown below.



[Figure 1]

Solution:

a) To draw the ray diagram, we'll follow the steps laid out above for converging lenses.

First we draw the a ray that travels parallel to the principle axis and refracts through the focus on the other side (the red ray). Next we draw a ray through the focus on the same side that refracts out parallel (the green ray). Finally we draw the ray that travels straight through the center of the lens without refracting (the blue ray). The result is shown below.



[Figure 2]

b) Based on the ray diagram and the initial position of the object, we know that the image is a real, inverted, and smaller than the original object.

c) To calculate the exact position of the object, we can use the lens maker's equation.

1ffuv1vv=1v−1u=2 units=−5 units=?=1f−1u=3.33 units

Time for Practice

1. Consider a converging lens with a focal length equal to three units and an object placed outside the focal point. 

Answer 2

• What is the formula for magnification obtained with a lens?

(a) Ratio of height of image to height of object $\huge\red{✓}$

(b) Double the focal length.

(c) Inverse of the radius of curvature.

(d) Inverse of the object distance.