Question

A convex lens of focal length 20 cm can produce a magnified virtual image as well as real image. Is this a correct statement? If true=1 and false=0.
Type 0 or 1 as your answer.

Answer 1

Use the formula for magnification for lens i.e. $m=\dfrac{v}{u}$ to check whether the image formed is magnified. Then use the formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ to check whether it forms a virtual as well as real image. Here, u is negative and f is positive.

Formula used:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$m=\dfrac{v}{u}$

Complete step by step answer:
To answer this question let us use the formula, which gives us the relation between the positions of the object and its image when the light from the object passes through a thin lens.
i.e. $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ ….. (i).
In this formula, u is the position of the object, v is the position of the image and f is the focal length of the lens. These values are according to the sign convection.
We will also use the formula of magnification of the image for a lens.
i.e. $m=\dfrac{v}{u}$ …. (ii), where m is the ratio of the height of the image to the height of the object.
Now consider a convex lens in front of an object. According to equation (ii), if v is greater than u then the m will be greater than 1. Hence, the image will be magnified.
Therefore, a convex lens can form a magnified image. Now we have to check whether it can form a magnified virtual as well as real image.
For this, we will use equation (i).
Before that let us understand when a real and virtual image is formed.
A real image is formed when the image is formed on the other side of the lens. i.e when v is positive.
A virtual image is formed when the image is formed on the same side of the lens where the rays are incident. i.e when v is negative.
According to the sign convection, the value of u is always negative. Let the distance of the object from the optical centre of the lens be d. hence, u = -d.
For a convex lens, focal length is always positive. Let it be f.
Substitute the values of u and f in equation (i).
Hence, we get
$\dfrac{1}{v}-\dfrac{1}{-d}=\dfrac{1}{f}$
$\Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{d}$
$\Rightarrow \dfrac{1}{v}=\dfrac{d-f}{fd}$
$\Rightarrow v=\dfrac{fd}{d-f}$ …. (iii)
In equation (iii), d and f are positive. The value of fd will also be positive because the product of two positive numbers is positive. However, the value of d-f may be positive or negative depending on the value of u only since f is a constant value.
If d > f, then d-f will be positive.
Then v will be the ratio of a positive number and a positive number. Hence, the value of v will also be positive.
This means that the image formed is a real image.
If d < f, then d-f will be negative.
Then v will be the ratio of a positive number and a negative number. Hence, the value of v will also be negative.
This means that the image formed is a virtual image.
Therefore, a convex lens can form a magnified virtual as well as real image depending on the position of the object. Which means that the given statement is true.

Hence, the correct answer is 1.

Note:
Note that the magnified real image is always inverted. This can be explained with the formula $m=\dfrac{v}{u}$.
We know that u is always negative and when a real image is formed v is positive. Hence, the value of m is negative.
When m is a negative value, the image formed is inverted.