Answer:
θ= 128.896°
Step-by-step explanation:
In the Given question
f= 950 MHz
width of slit =35 cm = 0.35m
the expression to calculate angular width
![sin(\theta)/(2)=(\lambda )/(width of slit)](https://img.qammunity.org/2020/formulas/physics/college/qpc8ru360jcgp8c2mhixqf58o0csbs01fv.png)
wavelength is
![\lambda =(c)/(f)](https://img.qammunity.org/2020/formulas/chemistry/middle-school/4lb4wq4s1obalscnua9guidwlffw097bwe.png)
![\lambda =(3* 10^8)/(950*10^6)](https://img.qammunity.org/2020/formulas/physics/college/gq7q7kylakchbjo57f4nlzw5ztdld53dwt.png)
λ= 0.315 m
therefore angular width
![sin(\theta)/(2)=(0.315 )/(0.35)](https://img.qammunity.org/2020/formulas/physics/college/4g6fsy86wv087zcku46qihzof76j1s5sfw.png)
![sin(\theta)/(2)= 0.9022](https://img.qammunity.org/2020/formulas/physics/college/ybbx5jxjknhp60ecolpsb8s1u20d6noq9x.png)
on further solving we get
θ= 2*64.448
θ= 128.896°
hence the horizontal angular width θ= 128.896°