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Comprehensive and Detailed Explanation (150–250 words):

Use the thin lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}. With f=+10 cmf = +10\,\text{cm} and do=30 cmd_o = 30\,\text{cm}: 1/di=1/10−1/30=(3−1)/30=2/30=1/151/d_i = 1/10 - 1/30 = (3 - 1)/30 = 2/30 = 1/15, so di=+15 cmd_i = +15\,\text{cm}. Positive image distance indicates a real image on the side opposite the object. The transverse magnification is m=−di/do=−15/30=−0.5m = -d_i/d_o = -15/30 = -0.5; the negative sign denotes inversion and the magnitude indicates the image is half the object’s size. Option B incorrectly claims an upright real image (real images from single converging lenses are inverted for objects outside the focal length) and wrong magnification. Options C and D describe virtual images (negative did_i) that would occur only if the object were inside the focal length (do