Question

A concave lens refracts parallel rays in such a way that they are bent away from the axis of the lens. For this reason, a concave lens is referred to as a diverging lens.

A) Consider the following diagrams, where F represents the focal point of a concave lens. In these diagrams, the image formed by the lens is obtained using the ray tracing technique. Which diagrams are accurate?

B) If the focal length of the concave lens is -7.50cm , at what distance from the lens should an object be placed so that its image is formed 3.70cm from the lens?

C) What is the magnification produced by the concave lens described in Part B?

d) Where should the object be moved to have a larger magnification?

The object should be moved closer to the lens.
The object should be moved farther from the lens.
The object should be moved to the focal point of the lens.
The object should not be moved closer to the lens than the focal point.

Answer 1

The concepts used to solve this problem are ray tracing and image formation with a concave lens.

First, find the accurate diagram using the refraction rules for a diverging lens.

Find the image position for diverging lens using the lens formula.

And also find magnification with object distance and image distance relation.

Finally, find the correct option for an object to have a larger magnification using the image characteristics of the concave lens.

Concave lenses are thin at the edges. So the ray passing through the lens will get diverged.

The concave lens is called diverging because it refracts parallel rays away from the axis of the lens.

The refracted rays diverge, and they appear to come from one point called the principal focus.

There are three refraction rules for the diverging lens,

The first rule is when any ray travels parallel to the principal axis it refracts through the lens and travels through the focal point.

The second rule is when a ray travels towards the focal point to the lens it will refract and travel parallel to the principal axis.

The third rule is when a ray passes through the center of the lens it will continue in the same direction.

The focal length is negative for the concave lens. So the image of an object should fall on the same side of the object in a concave lens.

Depending on the position of the object relative to the lens and its focal point, the lens produces different types of images.

The object from its pole is called object distance and the distance from the pole of the lens is called image distance.

The distance of the principal focus from the pole is called the focal length.

By using this object distance, image distance, and principal focus a new relationship is named as lens formula.

The expression for the lens formula is,

                                                               

Here, v is the object distance, u is the image distance and f is the focal length.

Image of an object is magnified with respect to the object size.

It is expressed as the ratio of the height of the image to the height of the object.

Here, u is the image distance and v is the object distance.

If the object is placed between the center of curvature and focal point the image will be formed beyond the center of curvature.

The absolute value of magnification will be graeter than one.

When moving the object near to the lens the magnification will be larger. And the image will be real.

(A)

Part A.1

The wrong diagrams are,

According to the rules of refraction for diverging lens, the focal length is negative for concave lens. So, the image of an object should fall on the same side of the object in concave lens.

But for these options B, D the image is on the opposite side of the object. These diagrams are not applicable.

The correct diagram is,

According to the rules of refraction for diverging lens,

Option A satisfies the third rule. That is, when a ray passes through the center of the lens it will continue in same direction.

Since option A is identified as applicable options.

(A)

(B)

The expression for the lens formula is,

The image is forming on the same object side, so the image distance should be in negative.

Substitute -3.70cm  for u and -7.50cm for f.

(C)

The expression for the magnification is,

Substitute 7.30cm for v and -3.70cm for u.

(D)

The wrong options are,

• The object should be moved farther from the lens.

• The object should be moved to the focal point of the lens.

• The object should not be moved closer to the lens less than the focal point.

No image will form, when the object is at focal point. When the object is moved closer to the lens less than the focal point, the image distance and image height approaches to infinity and when the object distance is zero, the image distance will be zero. Image formation is not real. So, these options are not applicable.

(D.1)

The correct option is that the object should be moved closer to the lens.

When the object is moved closer to the lens the image will become larger and away.

So, the correct option is identified.