Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
Answer:
T = reading (cm) time base (s / cm)
f = 1 / T
Explanation:
An oscilloscope is a piece of equipment that allows you to visualize and measure a wave that reaches you, in the case of having a sonometer this transforms the sound wave into an electrical signal to be introduced through one of the voltage channels of the equipment, on the screen we will see the oscillating alternating signal, if it is fixed we can make the reading, if it is moving the time base and the trigger must be adjusted to stop it.
In the oscilloscope we can read the period of the signal, this is the time it takes for the signal to repeat itself with this value, we can calculate the frequency with the formula, for the reading of the period the distance is measured on the labeled screen and multiplied by the time base
T = reading (cm) time base (s / cm)
f = 1 / T
Answer:
Explanation:
The energy of a photon is given by the equation 
, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of 
.
We also know that frequency and wavelength are related by 
, so we have 
, where c is the <em>speed of light</em>.
We will want the number of photons, so we can write
We need to know then how much energy do we have to calculate N. The equation of power is 
, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be 
.
Putting this paragraph in equations:
.
And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
