To solve for the focal length of a convex lens when the object and image distances are both equal to \(x\), we use the lens formula:
\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
where:
\(f\) = focal length of the lens
\(v\) = image distance
\(u\) = object distance
Given that \(v = x\) and \(u = x\), we substitute these values into the lens formula:
\[\frac{1}{f} = \frac{1}{x} + \frac{1}{x}\]
\[\frac{1}{f} = \frac{2}{x}\]
Taking the reciprocal to solve for \(f\):
\[f = \frac{x}{2}\]
Thus, the focal length of the convex lens is \(\frac{x}{2}\).
Let the object distance be u = –x (negative as it's on the left of the lens), and the image distance be v = +x.
Using the lens formula: 1/f = 1/v – 1/u
Substituting the values: 1/f = 1/x – (–1/x) = 1/x + 1/x = 2/x
So, the focal length f = x/2.
Final Answer: x/2
Assertion : Out of Infrared and radio waves, the radio waves show more diffraction effect.
Reason (R): Radio waves have greater frequency than infrared waves.