Question

A certain lens focuses light from an object 1.85 m away as an image 48.8 cm on the otherside of the lens. What type of lens is it?

diverging lens

converging lens
What is its focal length?
cm
Is theimage real or virtual?

real

virtual

Answer 1

Concepts and reason

The concepts used to solve this problem are thin lens formula, object distance, image distance, and focal length.

First, use the side of the image produced by the lens to determine the nature of the lens.

Use the thin lens formula that relates the object distance, image distance, and focal length of a lens to determine the focal length.

Finally, use the condition for the image to be real and determine the image produced by the lens.

Fundamentals

The expression for the thin lens formula is as follows:

$\frac{1}{f} = \frac{1}{i} + \frac{1}{o}$

Here, the focal length is $f$, the image distance is $i$, and the object distance is $o$.

Object distance is the distance between the object and the lens considered.

Image distance is the distance between the images formed by the lens and to the lens system.

Focal length is the distance between the lens and the point at which the collimated rays from an object are focused.

(a)

The converging lens focuses at all the collimated rays that are falling on them from one side to a point on the other side. Hence, the image is formed on the opposite side of the object for converging lens.

The incorrect option is

• Diverging lens

The diverging lens focuses at all the collimated rays that are falling on them from one side to a point on the same side. This occurs because the diverging lens will diverge all the rays. Hence, the image is formed on the same side of the object for convex lens.

The image is formed on the same side which implies that the lens is converging.

Therefore, the correct option is given below

• Converging lens

(b)

The expression for the thin lens formula is as follows:

$\frac{1}{f} = \frac{1}{i} + \frac{1}{o}$

Substitute $1.85\,{\rm{m}}$ for $o$, and $48.8\,{\rm{cm}}$ for $i$.

$\begin{array}{c}\\\frac{1}{f} = \frac{1}{{1.85\,{\rm{m}}}} + \frac{1}{{48.8\,{\rm{cm}}}}\\\\ = \frac{1}{{1.85\,{\rm{m}}}} + \frac{1}{{\left( {48.8\,{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{{{10}^2}\,{\rm{cm}}}}} \right)}}\\\\ = \frac{1}{{1.85\,{\rm{m}}}} + \frac{1}{{48.8 \times {{10}^{ - 2}}\,{\rm{m}}}}\\\\ = 2.59\,{\rm{m}}\\\end{array}$

Rearranging the above expression for $f$,

$\begin{array}{c}\\f = \frac{1}{{2.59\,{\rm{m}}}}\\\\ = 0.386\,{\rm{m}}\\\\{\rm{ = 38}}{\rm{.6}}\,{\rm{cm}}\\\end{array}$

The focal length of the converging lens is $38.6\,{\rm{cm}}$.

(c)

The condition for the converging lens to produce a real image is that the object distance should be greater than that of focal length. If the object distance is less than the focal length, the image produced by the lens will be a virtual image.

The incorrect option is given below:

• Virtual

The focal length of this converging lens is $36.8\,{\rm{cm}}$. The object distance is $1.85\,{\rm{m}}$.

Therefore, the image produced is real because the distance of the image is positive.

Therefore, the correct option is given below

• Real

Ans: Part a

The lens is converging lens.

Part b

The focal length of the lens is ${\bf{38}}{\bf{.6}}\,{\bf{cm}}$.

Part c

The image produced by the converging lens is real.