To solve the problem, we can use Snell's Law, which states that:
\[ \frac{\sin(\theta_r)}{\sin(\theta_i)} = \frac{v_r}{v_i} = \frac{\lambda_r}{\lambda_i} \]
where:
- \(\lambda_r\) is the wavelength in the new medium,
- \(\lambda_i\) is the wavelength in the original medium,
- \(v_r\) and \(v_i\) are the speeds of wave in the new and original mediums, respectively.
You are given:
- \(\frac{\sin(\theta_r)}{\sin(\theta_i)} = 0.816\)
- \(\lambda_r = 4.0 \times 10^{-7} , \text{m}\)
Using Snell's Law, we can relate the wavelengths:
\[ \frac{\lambda_r}{\lambda_i} = 0.816 \]
Rearranging the equation to find \(\lambda_i\):
\[ \lambda_i = \frac{\lambda_r}{0.816} \]
Substituting the value of \(\lambda_r\):
\[ \lambda_i = \frac{4.0 \times 10^{-7} , \text{m}}{0.816} \]
Calculating this gives:
\[ \lambda_i \approx 4.9 \times 10^{-7} , \text{m} \]
Thus, the wavelength in the original medium is approximately:
\[ \lambda_i \approx 4.9 \times 10^{-7} , \text{m} \]
So, the correct answer from the choices provided is:
\(\boxed{4.9 \times 10^{-7} , \text{m}}\)
1 answer