solve this---**Question Stem:** An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the sizes, of the images formed are equal, the focal length of the lens will be **Options:** a. 15 cm b. 17 cm c. 21 cm d. 11 cm

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Let's solve this lens problem step by step. We have an object that forms a real image when placed 16 centimeters from a lens, and a virtual image when placed 6 centimeters away. Both images have equal sizes. We need to find the focal length of the lens. Now let's set up the problem using the lens formula. The lens formula relates object distance u, image distance v, and focal length f. We use the sign convention where object distances are negative. For the first case, u₁ equals negative 16 centimeters, and for the second case, u₂ equals negative 6 centimeters. Let's apply the lens formula to both cases. For case 1 with object at negative 16 centimeters, we get v₁ equals 16f divided by 16 minus f. For case 2 with object at negative 6 centimeters, we get v₂ equals 6f divided by 6 minus f. Notice that v₁ is positive for a real image, while v₂ is negative for a virtual image. Since both images have equal sizes, we can set up the equation. The magnification for the first case gives us v₁ over 16, and for the second case gives us negative v₂ over 6. Setting these equal and substituting our expressions for v₁ and v₂, we can solve for f. After simplification, we get f equals 11 centimeters. Let's verify our answer. With f equals 11 centimeters, for the first case where the object is at 16 centimeters, we have 16 greater than 11, which satisfies the condition for a real image. For the second case at 6 centimeters, we have 6 less than 11, which satisfies the condition for a virtual image. Therefore, the focal length is 11 centimeters, which corresponds to option d.