<ol>
<li>If v is constant, how does f change if λ quadruples?</li>
<li>c is equal to 3x10^8 m/s. What is the value of n if v equals 2.25 x 10^8 m/s</li>
<li>If n2 is greater than n1, is Ө1 greater than, less than, or equal to Ө2? Prove using third equation.</li>
<li>Assuming Ө2 is 90 degrees, write an expression for Ө1 in terms of n1 and n2. </li>
</ol>
<ol>
<li>If v is constant, then f has to vary inversely with λ to counterbalance it. Therefore, if λ quadruples, f is quartered. </li>
<li>divide c by v. You have both variables there. </li>
<li>Set up a ratio. n1/n2 = sinӨ2/sinӨ2.</li>
<li>Use the third equation. sin90 is 1. Therefore, divide both sides by n1 and you get sinӨ1 = n2/n1. You want Ө1, so you have to take the arcsin of both sides and voila!</li>
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<p>but just an-add: notice the relations b/w the variables in an equation, most high school physics involve basic plug+ chug problems and basic interpretations of equations
… keep this in mind and you should be able to tackle physics and don’t give up!</p>
<p>You have n2/n1=sin(Ө1)/sin(Ө2). So sin(Ө1)>sin(Ө2).</p>
<p>Because sine is a very tricky function to deal with regarding greater than/less than, you’ll have to slog through a whole slew of cases. If you want a short answer, though, Ө1 can be greater than Ө2 and vice versa. More details are needed to completely solve the problem.</p>
<p>You have n2/n1=sin(Ө1)/sin(Ө2). So sin(Ө1)>sin(Ө2).</p>
<p>Because sine is a very tricky function to deal with regarding greater than/less than, you’ll have to slog through a whole slew of cases. If you want a short answer, though, Ө1 can be greater than Ө2 and vice versa. More details are needed to completely solve the problem. "</p>
<p>In this problem, we are restricted to 0 < theta < 90 so it is fairly simple. When the angle is in that interval, sine goes up as the angle goes up. </p>
<p>n2/n1 > 1 means sin(theta1)/sin(theta2) > 1 so sin(theta1) > sin(theta2) and then theta1>theta2</p>