An object is located on the optical axis of a biconvex thin lens at the distance S_(1)=20 cm from its optical center. The focal length of the lens is f=10 10 cm. What is the distance S_(2) between the lens and the image? Is the image real or virtual? Please select one answer option S_(2)=30cm the image is virtual S_(2)=30 cm, the image is real S_(2)=10 cm, the image is virtual S_(2)=20 cm, the image is real

To solve this problem, we can use the thin lens formula:<br /><br />\[<br />\frac{1}{f} = \frac{1}{S_1} + \frac{1}{S_2}<br />\]<br /><br />where \( f \) is the focal length of the lens, \( S_1 \) is the object distance, and \( S_2 \) is the image distance.<br /><br />Given:<br />- \( f = 10 \) cm<br />- \( S_1 = 20 \) cm<br /><br />We need to find \( S_2 \).<br /><br />First, rearrange the thin lens formula to solve for \( S_2 \):<br /><br />\[<br />\frac{1}{S_2} = \frac{1}{f} - \frac{1}{S_1}<br />\]<br /><br />Substitute the given values:<br /><br />\[<br />\frac{1}{S_2} = \frac{1}{10} - \frac{1}{20}<br />\]<br /><br />Find a common denominator to combine the fractions:<br /><br />\[<br />\frac{1}{S_2} = \frac{2}{20} - \frac{1}{20} = \frac{1}{20}<br />\]<br /><br />So,<br /><br />\[<br />S_2 = 20 \text{ cm}<br />\]<br /><br />Since \( S_2 \) is positive, the image is real.<br /><br />Therefore, the correct answer is:<br /><br />\( S_2 = 20 \) cm, the image is real.