real image of equal size is obtained at listance of 48 cm from the lens. The type of th ens and its focal length is : o) convex lens of focal length 24 cm b) convex lens of focal length 48 cm c) concave lens of focal length of 48 cm d) concave lens of focal length 24 cm

To determine the type of lens and its focal length, we can use the lens equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where: - \( f \) is the focal length of the lens, - \( d_o \) is the object distance (48 cm in this case), - \( d_i \) is the image distance (which we need to find). Given that the image is real and of equal size, we can assume that the magnification is 1. This means that the image distance is equal to the object distance, i.e., \( d_i = 48 \) cm. Now, we can substitute the values into the lens equation to find the focal length of the lens. Solve the equation \( \frac{1}{f} = \frac{1}{48} + \frac{1}{48} \). Solve the equation by following steps: - step0: Solve for \(f\): \(\frac{1}{f}=\frac{1}{48}+\frac{1}{48}\) - step1: Find the domain: \(\frac{1}{f}=\frac{1}{48}+\frac{1}{48},f\neq 0\) - step2: Add the terms: \(\frac{1}{f}=\frac{1}{24}\) - step3: Rewrite the expression: \(f=24\) - step4: Check if the solution is in the defined range: \(f=24,f\neq 0\) - step5: Find the intersection: \(f=24\) The focal length of the lens is 24 cm. Since the focal length is positive, the lens is a convex lens. Therefore, the correct answer is: a) Convex lens of focal length 24 cm.