Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Communication Systems
0 votes

The AM radio wave bends appreciably round the corners of a $1m \times 1m$ board but the FM wave only negligibly bands. If the average wavelength of the AM and FM waves are $ \lambda_a$ and $\lambda_f$ , then

$\begin {array} {1 1} (a)\;\lambda_a > \lambda_f & \quad (b)\;\lambda_a = \lambda_f \\ (c)\;\lambda_a < \lambda_f & \quad (d)\;\text{ can't be determined} \end {array}$


Can you answer this question?

1 Answer

0 votes
Bending of wave around the corner occurs, if the dimension of the board is comparable to the wavelength of the wave.
From the given information we know that $ \lambda_a$ since no bending of FM waves take place two possibilities may arise whether $ \lambda_f>>1m \: or\: \lambda_f<<1m$.
$ \lambda_f>>1m \: or\: \lambda_f<<1m$.
So, information is insufficient to determine whether $ \lambda_a>\lambda_f$ is comparable to $1m$. But, or $\lambda_a<\lambda_f.$
Ans : (d)
answered Feb 28, 2014 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App