Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,
[tex]f=\frac{c}{\lambda}[/tex]
substituting the values in the equation we get,
[tex]f=\frac{3.0\times 10^8 m/s}{2.9m}[/tex]
f = 1.03 x 10⁸Hz
Now,
The time period (T) = [tex]\frac{1}{f}[/tex]
or
T = [tex]\frac{1}{1.03\times 10^8}[/tex] = 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target = [tex]\frac{290}{2}[/tex] = 145 m