Let's break down the solution step by step. --- ## **Step 1: Understanding the Problem** - There are \(N\) layers. - The top layer has refractive index \(n\). - Each subsequent layer has a refractive index \(k\) times **smaller** than the one above: - 1st layer: \(n\) - 2nd layer: \(n/k\) - 3rd layer: \(n/k^2\) - ... - \(N\)th layer: \(n/k^{N-1}\) We want the **minimum angle of incidence** \(\alpha_{\min}\) such that **light does not exit the pile**—i.e., total internal reflection occurs at some point. --- ## **Step 2: Where Does Total Internal Reflection Happen?** Total internal reflection occurs when light travels from a **denser** (higher \(n\)) to a **rarer** (lower \(n\)) medium at an angle greater than the critical angle. The **lowest** refractive index is at the **bottom**, so the "most dangerous" interface (most likely to let light escape) is at the **bottom layer**, between \(n/k^{N-1}\) (bottom layer) and whatever is below (let's assume air, \(n_a \approx 1\)). --- ## **Step 3: Find the Critical Angle** The **critical angle** \(\theta_c\) from bottom layer to air is: \[ \sin \theta_c = \frac{n_a}{n/k^{N-1}} \] \[ \sin \theta_c = \frac{1}{n/k^{N-1}} = \frac{k^{N-1}}{n} \] So, \[ \theta_c = \arcsin\left(\frac{k^{N-1}}{n}\right) \] --- ## **Step 4: Relate Incident Angle to Critical Angle** Using **Snell's Law** at each interface: Let the angle in the top layer be \(\alpha\). At each interface: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] So, for the top surface (air to layer 1): \[ \sin \alpha = \frac{n}{1} \sin \theta_1 \implies \sin \theta_1 = \frac{\sin \alpha}{n} \] But we want to relate the **incident angle in the very first layer** to the **angle in the bottom layer**, since that's where total internal reflection matters. Let the **angle in the bottom layer** be \(\theta_N\). From repeated application of Snell's Law across layers: \[ n \sin \theta_1 = \frac{n}{k} \sin \theta_2 = \frac{n}{k^2} \sin \theta_3 = \cdots = \frac{n}{k^{N-1}} \sin \theta_N \] So, \[ n \sin \theta_1 = \frac{n}{k^{N-1}} \sin \theta_N \] \[ \sin \theta_N = k^{N-1} \sin \theta_1 \] From above, \(\sin \theta_1 = \frac{\sin \alpha}{n}\): \[ \sin \theta_N = k^{N-1} \cdot \frac{\sin \alpha}{n} \] --- ## **Step 5: Apply the Total Internal Reflection Condition** For total internal reflection at the **bottom layer**: \[ \theta_N \geq \theta_c \implies \sin \theta_N \geq \sin \theta_c = \frac{k^{N-1}}{n} \] So: \[ k^{N-1} \cdot \frac{\sin \alpha}{n} \geq \frac{k^{N-1}}{n} \] \[ \sin \alpha \geq 1 \] But \(\sin \alpha = 1\) only possible for \(\alpha = 90^\circ\), which is not practical. So let's double-check: the correct application is that, **for critical angle**, \(\sin \theta_N = \frac{k^{N-1}}{n}\), and that's the **maximum** possible before refraction occurs into air. We want to find the \(\alpha\) for which this is true. So, \[ k^{N-1} \cdot \frac{\sin \alpha}{n} = \frac{k^{N-1}}{n} \] \[ \sin \alpha = 1 \] \[ \alpha = 90^\circ \] But clearly, this suggests that for any real incidence angle, light cannot escape unless \(\alpha = 90^\circ\). So let's check the setup again. If the light must **not exit** the pile, we need the **angle in the bottom layer** to be **greater than or equal to the critical angle** for **all rays**. But the smallest \(\alpha\) for which this is true is when the angle in the bottom layer **just reaches** the critical angle. So, \[ \sin \theta_N = \frac{k^{N-1}}{n} \] But \[ \sin \theta_N = k^{N-1} \cdot \frac{\sin \alpha}{n} \] Set equal: \[ k^{N-1} \cdot \frac{\sin \alpha_{\min}}{n} = \frac{k^{N-1}}{n} \] \[ \sin \alpha_{\min} = 1 \] \[ \alpha_{\min} = 90^\circ \] --- ## **Final Answer (with Explanation)** ### **Minimum angle of incidence** \[ \boxed{90^\circ} \] ### **Explanation** - For the light to **not pass through the pile**, it must undergo **total internal reflection** at the bottommost interface. - The critical angle for the bottom interface is \(\theta_c = \arcsin \left(\frac{k^{N-1}}{n}\right)\). - Snell's Law across layers gives \(\sin \theta_N = k^{N-1} \frac{\sin \alpha}{n}\). - Setting \(\sin \theta_N = \sin \theta_c\), we find \(\sin \alpha_{\min} = 1\), or \(\alpha_{\min} = 90^\circ\). - **Physically, this means total internal reflection is only guaranteed for incident angles at or above \(90^\circ\), which is not possible, so for any real angle, some light may pass through unless the refractive index contrast is such that \(n \gg k^{N-1}\).** --- ### **Summary Table** | Layer | Refractive Index | |-----------------|-----------------| | Top (1st) | \(n\) | | 2nd | \(n/k\) | | 3rd | \(n/k^2\) | | ... | ... | | Bottom (\(N\)th)| \(n/k^{N-1}\) | --- If you need the answer in terms of a formula *if* the external medium has a higher index, or if you want a more general expression, please specify!