1.3. An airglow layer extends from 90 km to 110 km. The volume emission rate is zero at 90 km and increases linearly with increasing altitude to 75 × 10^6 photons m^−3 s^−1 at 100 km, then decreases linearly with increasing altitude to zero at 110 km. A photometer with a circular input 0.1m in diameter and a field of view of 1 degree half-angle views the layer at an angle of 45 degrees above the horizon.

*Problem:*1.3. An airglow layer extends from 90 km to 110 km. The volume emission rate is zero at 90 km and increases linearly with increasing altitude to 75 × 10^6 photons m^−3 s^−1 at 100 km, then decreases linearly with increasing altitude to zero at 110 km. A photometer with a circular input 0.1m in diameter and a field of view of 1 degree half-angle views the layer at an angle of 45 degrees above the horizon.

(a) Determine the vertically integrated emission rate in Rayleigh.

(b) Calculate the vertically viewed radiance of the layer in photon units.

(c) Calculate the vertically viewed radiance of the layer in energy units, for a wavelength of 557.7 nm.

(d) Calculate the photon rate into the instrument.

**Solution/Hints/Data:**

Here we know following according to the problem:

**V(90 km) = 0, we can consider it as r1,***Where V is the volume emission rate in photons m^−3 s^−1***V(100 km) = 75 x 10^6, we can consider it as r2****Area =****πr^2,***where, r=0.05 m***Half Angle= 1 degree**

*The only problem I am facing is how to integrate V(r) along “r” according to the following formula as how can we integrate a function V(r) here if we don’t know on which variables it is dependent.*

*Following is the concerned text I am pasting here from the source book relating to the formula for describing relationship between integrated emission rate and emitting layers:*