Let's solve **both questions step by step** using the **Lens Maker's Formula**: \[ \frac{1}{f} = (n - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right] \] where: - \( f \) = focal length - \( n \) = refractive index of material - \( R_1 \) = radius of curvature of first surface (positive for convex, negative for concave) - \( R_2 \) = radius of curvature of second surface (positive if the center of curvature is to the right of the surface) --- ### **First Question:** **One surface is convex with \( R_1 = +50 \) cm.** #### **Step 1: Write the Lens Maker's Formula with values** Given: - \( f = 100 \) cm - \( n = 2.4 \) - \( R_1 = +50 \) cm - \( R_2 = ? \) \[ \frac{1}{100} = (2.4 - 1) \left[ \frac{1}{50} - \frac{1}{R_2} \right] \] **Explanation:** Substituted given values into the lens maker's formula. --- #### **Step 2: Simplify the Equation** \[ \frac{1}{100} = 1.4 \left[ \frac{1}{50} - \frac{1}{R_2} \right] \] \[ \frac{1}{100} \div 1.4 = \frac{1}{50} - \frac{1}{R_2} \] \[ \frac{1}{140} = \frac{1}{50} - \frac{1}{R_2} \] **Explanation:** Divided both sides by 1.4 to isolate the bracket. --- #### **Step 3: Solve for \( R_2 \)** \[ \frac{1}{R_2} = \frac{1}{50} - \frac{1}{140} \] \[ \frac{1}{R_2} = \frac{140 - 50}{7000} \] \[ \frac{1}{R_2} = \frac{90}{7000} = \frac{9}{700} \] \[ R_2 = \frac{700}{9} \approx 77.8 \text{ cm} \] **Explanation:** Found the value for \( R_2 \) by calculating the difference. --- ### **Final Answer (First Question):** **The radius of curvature of the other surface should be approximately \( \boxed{77.8\ \text{cm}} \).** --- --- ### **Second Question:** **One surface is concave with \( R_1 = -50 \) cm.** #### **Step 1: Write the Lens Maker's Formula with values** Given: - \( f = 100 \) cm - \( n = 2.4 \) - \( R_1 = -50 \) cm - \( R_2 = ? \) \[ \frac{1}{100} = (2.4 - 1) \left[ \frac{1}{-50} - \frac{1}{R_2} \right] \] **Explanation:** Substituted given values into the formula. --- #### **Step 2: Simplify the Equation** \[ \frac{1}{100} = 1.4 \left[ -\frac{1}{50} - \frac{1}{R_2} \right] \] \[ \frac{1}{100} \div 1.4 = -\frac{1}{50} - \frac{1}{R_2} \] \[ \frac{1}{140} = -\frac{1}{50} - \frac{1}{R_2} \] \[ \frac{1}{R_2} = -\frac{1}{50} - \frac{1}{140} \] **Explanation:** Distributed the negative sign and isolated \(\frac{1}{R_2}\). --- #### **Step 3: Solve for \( R_2 \)** \[ \frac{1}{R_2} = -\left( \frac{1}{50} + \frac{1}{140} \right) \] \[ \frac{1}{R_2} = -\left( \frac{140 + 50}{7000} \right) \] \[ \frac{1}{R_2} = -\frac{190}{7000} \] \[ R_2 = -\frac{7000}{190} \approx -36.8 \text{ cm} \] **Explanation:** Solved for \( R_2 \) as a negative value (which means the center of curvature is on the same side as the incoming light). --- ### **Final Answer (Second Question):** **The radius of curvature of the other surface should be approximately \( \boxed{-36.8\ \text{cm}} \).**