Question

The focal length of a convex lens is f. An object is placed at a distance x from its first focal point. The ratio of the size of the real image to that of the object is
A) $ \dfrac{f}{{{x^2}}} $
B) $ \dfrac{{{x^2}}}{f} $
C) $ \dfrac{x}{f} $
D) $ \dfrac{f}{x} $

Answer 1

Hint: Here, firstly we find the size of real image by using lens formula i.e. $ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} $ , v is the size of image and u is the distance of object and f is focal length of the convex lens. Now it is given that the object is placed at a distance x from its first focal point so $ u = f + x $ , on substituting the values we will get the value of the size of real image i.e. v after that we can find the ratio of v and u.

Complete Step by step solution
Here, it is given that, Focal length of convex lens = f
As the object is placed at a distance x from its first focal point, so the distance of object is $ u = f + x $
Let v is the distance or size of the real image
Now, using the lens formula i.e.
 $ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} $ …………………… (1)
Where, f is the focal length of lens
U is the size or distance of object
V is the size or distance of the real image
Now, substitute the value of u in equation (1), we get
 $
   \Rightarrow \dfrac{1}{v} + \dfrac{1}{{f + x}} = \dfrac{1}{f} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{{f + x}} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{{f + x - f}}{{f\left( {f + x} \right)}} = \dfrac{x}{{f\left( {f + x} \right)}} \\
   \Rightarrow v = \dfrac{{f\left( {f + x} \right)}}{x} \\
  $
Therefore, the size of the real image is $ v = \dfrac{{f\left( {f + x} \right)}}{x} $ and the size of the object is $ u = f + x $ , thus we can find the ratio of size of real image to the size of the object i.e.
 $
   \Rightarrow \dfrac{v}{u} = \dfrac{{\dfrac{{f\left( {f + x} \right)}}{x}}}{{f + x}} \\
   \Rightarrow \dfrac{v}{u} = \dfrac{f}{x} \\
  $
Hence, the ratio of the size of real image to the size of object is $ \dfrac{f}{x} $
Therefore, option D is correct.

Note:
Here, it must be notice that we have to find the ratio of the ratio of the size of the real image to that of the object it means we have to find out the value of v/u
One thing should also be notice that the magnification of image is defined as the ratio of the size of image to the size of object, it means we have to find out the value of magnification i.e. $ m = \dfrac{v}{u} $ , so by directly putting the value of v and u we get the value of magnification or we can also say that for above question the value of magnification is $ m = \dfrac{f}{x} $ .