QuestionJuly 17, 2025

Laser light of wavelength 510. nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n=1.30 The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero)thickness of the coating for which transmission into the rod is minimized? Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: square times 10^square square Answer units

Laser light of wavelength 510. nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n=1.30 The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero)thickness of the coating for which transmission into the rod is minimized? Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: square times 10^square square Answer units
Laser light of wavelength 510. nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has n=1.30 The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero)thickness of the coating for which transmission into the rod is minimized? Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: square times 10^square square Answer units

Solution
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Answer

77.27 \text{ nm} Explanation 1. Identify the condition for destructive interference For minimum transmission (destructive interference), the optical path difference should be (m + \frac{1}{2})\lambda, where m is an integer. 2. Calculate the wavelength in the coating The wavelength in the coating is given by \lambda_c = \frac{\lambda_0}{n_c}, where \lambda_0 = 510 \text{ nm} and n_c = 1.65. So, \lambda_c = \frac{510 \text{ nm}}{1.65}. 3. Determine the thickness for first minimum For the first minimum (m=0), the thickness t of the coating is \frac{\lambda_c}{4}. 4. Calculate the thickness Substitute \lambda_c from Step 2 into the formula: t = \frac{510 \text{ nm}}{4 \times 1.65}.

Explanation

1. Identify the condition for destructive interference<br /> For minimum transmission (destructive interference), the optical path difference should be $(m + \frac{1}{2})\lambda$, where $m$ is an integer.<br /><br />2. Calculate the wavelength in the coating<br /> The wavelength in the coating is given by $\lambda_c = \frac{\lambda_0}{n_c}$, where $\lambda_0 = 510 \text{ nm}$ and $n_c = 1.65$. So, $\lambda_c = \frac{510 \text{ nm}}{1.65}$.<br /><br />3. Determine the thickness for first minimum<br /> For the first minimum ($m=0$), the thickness $t$ of the coating is $\frac{\lambda_c}{4}$. <br /><br />4. Calculate the thickness<br /> Substitute $\lambda_c$ from Step 2 into the formula: $t = \frac{510 \text{ nm}}{4 \times 1.65}$.
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