Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
Answer:
6.83 cm
Explanation:
Given data:
Focal length of the eye piece, = 2.50 cm
Focal length of the converging lens, f = 1.00 cm
distance of the object, p = 1.30 cm
Now,
we have the lens equation as:
q is the distance of the image
thus,
on substituting the values in the above equation, we get
or
q = 4.33 cm
now, the image is formed at the focal point of the eye piece,
therefore, the distance between the objective and the eyepiece, d = + q = 2.50 cm + 4.33 cm
or
d = 6.83 cm