Question

A real image is formed by a convex lens. When it is cemented with a concave lens, again the real image is formed. The real image formed by concave lens:
A. Shifts towards the lens system
B. Shifts away from the lens system
C. Remain in its original position
D. Shifts to infinity

Answer 1

Hint: Concave lens and convex lens are a type of lens. A lens is defined as an optical device that focuses or disperse a light beam by means of refraction. A convex lens forms a real image whereas a concave lens forms a virtual image.

Complete step by step answer:
A concave lens is a lens that diverges a light beam coming from the source into a diminished, upright, and virtual image. On the other hand, a convex lens is a lens that converges a light beam coming from the source into a real and inverted image. In the given question, a convex lens is placed in contact with the concave lens, therefore, the focal length of the combination is given by,
${\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{\left(-f_2\right)}}$
Here, $F$ is the focal length of the combination, $f_1$ is the focal length of the convex lens, and $f_2$ is the focal length of the concave lens.

From the above equation, we can say that the focal length of the combination is positive. Therefore, we can say that the combination of lenses is a convex lens.
Now, using the lens formula, we get
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{-u}}={\displaystyle \frac{1}{F}}$
$\Rightarrow {\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{F}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{F}}-{\displaystyle \frac{1}{u}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{u-F}{uF}}$
$\Rightarrow v={\displaystyle \frac{uF}{u-F}}$
Now, dividing the numerator and denominator on the Right-hand side by $F$ , we get
$v={\displaystyle \frac{u}{{\displaystyle \frac{u}{F}}-1}}$
Now, as the focal length of the combination $F$ is increased, therefore, ${\displaystyle \frac{u}{F}}$ decreases. With the decrease in ${\displaystyle \frac{u}{F}}$ , the distance of the image from the lens $v$ will increase. Therefore, the image will be far away from the lens. Therefore, the real image formed by a concave lens shifts far away from the lens system.

Hence, option B is correct.

Note: From the above solution, we can say that the distance of the image depends upon the focal length.But in simple words, we can say that a concave lens diverges the incident ray. Therefore, when the concave lens is placed in contact with the convex lens, the real image will be shifted away from the lens.