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Answer:
Explanation:
Let the angle between the first polariser and the second polariser axis is θ.
By using of law of Malus
(a)
Let the intensity of light coming out from the first polariser is I'
.... (1)
Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.
By the law of Malus
![I'' = I'Cos^{2}\left ( 90-\theta \right )](https://tex.z-dn.net/?f=I%27%27%20%3D%20I%27Cos%5E%7B2%7D%5Cleft%20%28%2090-%5Ctheta%20%5Cright%20%29)
So,
![I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )](https://tex.z-dn.net/?f=I%27%27%20%3D%20I_%7B0%7DCos%5E%7B2%7D%5Ctheta%20Cos%5E%7B2%7D%5Cleft%20%28%2090-%5Ctheta%20%5Cright%20%29)
![I'' = I_{0}Cos^{2}\theta Sin^{2}\theta](https://tex.z-dn.net/?f=I%27%27%20%3D%20I_%7B0%7DCos%5E%7B2%7D%5Ctheta%20Sin%5E%7B2%7D%5Ctheta)
![I'' = \frac{I_{0}}{4}Sin^{2}2\theta](https://tex.z-dn.net/?f=I%27%27%20%3D%20%5Cfrac%7BI_%7B0%7D%7D%7B4%7DSin%5E%7B2%7D2%5Ctheta)
(b)
Now differentiate with respect to θ.
![I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta](https://tex.z-dn.net/?f=I%27%27%20%3D%20%5Cfrac%7BI_%7B0%7D%7D%7B4%7D%5Ctimes%202%20%5Ctimes%202%20%5Ctimes%20Sin2%5Ctheta%20%5Ctimes%20Cos%202%5Ctheta)
![I'' = \frac{I_{0}}{2}\times Sin 4\theta](https://tex.z-dn.net/?f=I%27%27%20%3D%20%5Cfrac%7BI_%7B0%7D%7D%7B2%7D%5Ctimes%20Sin%204%5Ctheta)
Answer:
The correct option is;
a- sea surface temperature anomaly, in degrees Celsius
Explanation:
From the diagram related to the question we have two graphs super imposed of Sea surface temperature anomaly, in degrees Celsius and cholera incidence anomaly (%) both plotted against time in years.
On the left the y-axis represents the sea surface temperature anomaly while on the right, the y-axis represents the cholera incidence anomaly (%).
The display of the graph shows the sea surface temperature anomaly in blue.