(III) A bright object is placed on one side of a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance d_T = dᵢ + dₒ between the object and the screen is kept fixed, but the lens can be moved.

(a) Show that if d_T > 4ƒ, there will be two positions where the lens can be placed and a sharp image will be produced on the screen.

(b) If d_T < 4ƒ, show that there will be no lens position where a sharp image is formed.

(c) Determine a formula for the distance between the two lens positions in part (a), and the ratio of the image sizes.

Verified step by step guidance

1. Step 1: Recall the lens formula, which is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the lens, \( d_o \) is the distance from the lens to the object, and \( d_i \) is the distance from the lens to the image.
2. Step 2: Understand that the total distance between the object and the screen is fixed, \( d_T = d_o + d_i \). Substitute \( d_i = d_T - d_o \) into the lens formula to get a quadratic equation in terms of \( d_o \).
3. Step 3: Solve the quadratic equation obtained in Step 2 for \( d_o \). The discriminant of this quadratic equation will help determine the number of real solutions for \( d_o \), which correspond to the positions of the lens where a sharp image is formed.
4. Step 4: Analyze the discriminant. If \( d_T > 4f \), the discriminant is positive, indicating two real solutions for \( d_o \) and hence two positions for the lens where a sharp image is formed. If \( d_T < 4f \), the discriminant is negative, indicating no real solutions and hence no position where a sharp image is formed.
5. Step 5: For the case where \( d_T > 4f \) and there are two solutions for \( d_o \), calculate the distance between these two lens positions. Also, use the magnification formula \( m = -\frac{d_i}{d_o} \) to find the ratio of the image sizes at these two positions.