This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — The problem involves a concave spherical mirror with a focal length f = +10 cm. We will use the mirror formula and magnification formula to calculate the image distance and describe the image for different object distances. The mirror formula is: (1)/(f) = (1)/(d_o) + (1)/(d_i) where f is the focal length, d_o is the object distance, and d_i is the image distance. Rearranging for d_i: (1)/(d_i) = (1)/(f) - (1)/(d_o) d_i = (f · d_o)/(d_o - f) The magnification M is given by: M = -(d_i)/(d_o) If d_i > 0, the image is real. If d_i < 0, the image is virtual. If M < 0, the image is inverted. If M > 0, the image is upright. If |M| > 1, the image is magnified. If |M| < 1, the image is diminished. If |M| = 1, the image is the same size. Given: f = +10 cm a) Object distance d_o = 25 cm Step 1: Calculate the image distance d_i. (1)/(d_i) = (1)/(10 cm) - (1)/(25 cm) Find a common denominator (50): (1)/(d_i) = (5)/(50 cm) - (2)/(50 cm) (1)/(d_i) = (3)/(50 cm) d_i = (50)/(3) cm d_i ≈ 16.67 cm Step 2: Calculate the magnification M. M = -(d_i)/(d_o) = -50/3 cm25 cm M = -(50)/(3 × 25) = -(2)/(3) Step 3: Describe the image. Since d_i is positive, the image is real. Since M is negative, the image is inverted. Since |M| = 2/3 < 1, the image is diminished. The image distance is 16.67 cm. The image is real, inverted, and diminished. b) Object distance d_o = 10 cm Step 1: Calculate the image distance d_i. (1)/(d_i) = (1)/(10 cm) - (1)/(10 cm) (1)/(d_i) = 0 This implies d_i → . Step 2: Calculate the magnification M. M = -(d_i)/(d_o) = -()/(10 cm) = - Step 3: Describe the image. When the object is placed at the focal point of a concave mirror, the reflected rays are parallel and form an image at infinity. The image is real, inverted, and highly magnified (at infinity). The image distance is . The image is real, inverted, and highly magnified. c) Object distance d_o = 5 cm Step 1: Calculate the image distance d_i. (1)/(d_i) = (1)/(10 cm) - (1)/(5 cm) Find a common denominator (10): (1)/(d_i) = (1)/(10 cm) - (2)/(10 cm) (1)/(d_i) = -(1)/(10 cm) d_i = -10 cm Step 2: Calculate the magnification M. M = -(d_i)/(d_o) = --10 cm5 cm M = +2 Step 3: Describe the image. Since d_i is negative, the image is virtual. Since M is positive, the image is upright. Since |M| = 2 > 1, the image is magnified. The image distance is -10 cm. The image is virtual, upright, and magnified. Got more? Send 'em!