why am I getting matrix dimension error..?

when i tried to run this code , its displaying the "matrix dimension must agree". I want the plot for phase angle at various frequencies. and when i used for loop i got the result only for last iteration and plot shows only last iterated point. can i get a help on this.

clc;
clear;
close all;
d=0.23;
theta=40;
f=1:18;
lamda(f)=3*0.1/f;
phaseangle(lamda)=(2*pi/(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
%%%%%
clc;
clear;
close all;
d=0.23;
theta=40;
for f=1:1:18;
lamda=3*0.1/f;
phaseangle=(2*pi/(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
end

Answer 1

[Yongjian Feng]

There are several errors typos.

clc;
clear;
close all;
d=0.23;
theta=40;
f=1:18;
lambda =3*0.1./f;
phaseangle=(2*pi./(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
%%%%%
clc;
clear;
close all;
d=0.23;
theta=40;
for f=1:1:18
lambda=3*0.1/f;
phaseangle=(2*pi./(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
end

[Sai Monika Ananthoju]

thank you for your response.can you please help me with this.

[Yongjian Feng]

You actually want this, right?

clc;
clear;
close all;
d=0.23;
theta=40;
f=1:18;
lambda=3*0.1./f;
phaseangle=(2*pi./(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')

[Sai Monika Ananthoju]

Yes sir. Thank you.