Question

Plane microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of $14$ successive maxima, the detector travels a distance $0.14\text{m}.$ If the velocity of light is $3\times {10}^8{\text{ms}}^{-1}.$ The frequency of the transmitter is-

• A. $1.5\times {10}^{10}\text{Hz}$
• B. ${10}^{10}\text{Hz}$
• C. $2\times {10}^{10}\text{Hz}$
• D. $6\times {10}^{10}\text{Hz}$

Answer 1

The correct option is A $1.5\times {10}^{10}\text{Hz}$

Given,

Distance travelled by detector = $0.14\text{m}$
Speed of light, $c=3\times {10}^8{\text{ms}}^{-1}$


Distance between the two successive maxima $=\frac{\lambda }{2}$

According to question, distance travelled by detector in $14$ successive maxima

$\frac{14\lambda }{2}=0.14$

$\lambda =\frac{2}{100}\text{m}$

$\because \nu =\frac{c}{\lambda }$

$\Rightarrow \nu =\frac{3\times {10}^8\times 100}{2}=1.5\times {10}^{10}\text{Hz}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.