Question

An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the sizes, of the images formed are equal, the focal length of the lens will be
A.)15 cm
B.)17 cm
C.)21 cm
D.)11 cm

Answer 1

Hint: From given information, try to find out the type of lens used. Use the condition that the image formed is real or virtual, to determine the magnification of the lens. Then using the equation of the lens, we can get an appropriate solution.

Complete step-by-step answer:
It is given that both images formed are virtual and real. According to the characteristics of a lens, it must be a convex lens.

Let u be the distance of an object from a lens and v be the distance of image formed from a lens.

Also consider f be the focal length and m be the magnification of the lens.

We know that, the equation of lens as $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

v is not given so we convert the above equation in terms of magnification and focal length.
$1+\dfrac{u}{v}=\dfrac{u}{f}$
$1+\dfrac{1}{m}=\dfrac{u}{f}$

Since, magnification id the ratio of image distance to object distance.

$m=\dfrac{f}{f+u}$

For $u=-16cm$, the image formed is real. So, the value of $m=-1$. Putting this value in above equation we get,

$-1=\dfrac{f}{f-16}$
                                                                                         …..(1)
Now, for $u=-6cm$ , the image formed is virtual. So, the value of $m=1$. Putting this value in above equation, we get,
$1=\dfrac{f}{f-6}$ …..(2)
Solving equations (1) and (2) we get,
$f=11cm$
Correct option is D.

Note: Always remember, according to sign convention object distance is negative while image distance depends on if it is virtually formed or real. There is only one difference between the formula of mirror and formula of lens and it is of only the sign. In the mirror equation it is negative and in the lens equation, it is positive.