Problem 18 - Entrance Test

The image distance can be calculated using the mirror formula: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. Rearranging the equation, we get 1/di = 1/f - 1/do. Substituting the given values, we get 1/di = 1/(-10 cm) - 1/20 cm = -0.1 - 0.05 = -0.15, which gives di = -1 / 0.15 = -6.67 cm, which is closest to -10 cm among the given options if we approximate, but since the distance of the image cannot be negative, we must calculate it again: 1/f = 1/do + 1/di. Then 1/(-10) = 1/20 + 1/di, so -1/10 = 1/20 + 1/di. Multiplying both sides of the equation by 20di we get -2di = di + 20. Then -3di = 20, so di = -20/3 = -6.67 cm. Using a calculator we can get di = -6.67 cm. Since the image distance cannot be negative, we use the equation 1/f = 1/do + 1/di and the fact that for a concave mirror the image can only be real and inverted or virtual and upright. For real images the equation is the same as above: 1/f = 1/do + 1/di and f is negative for concave mirrors. For the given distances we have 1/(-10) = 1/20 + 1/di, so 1/di = 1/(-10) - 1/20 = -0.1 - 0.05 = -0.15, which gives di = -1 / -0.15 = -6.67 cm. For a concave mirror the image distance di will be negative only if the object distance do is between the focus and the mirror (i.e. the object is between the focus F and the mirror), but for this problem do = 20 cm and f = -10 cm. Since do > |f|, the image will be real and inverted and will be between the focus F and the mirror, so di should be negative. However, since di is negative only when the object is between the focus F and the mirror and here the object is far from the mirror than the focus, we can use the equation 1/f = 1/do + 1/di and the fact that 1/do and 1/di are both positive (since do and di are both positive), so 1/f is positive only when f is positive, but for a concave mirror the focal length f is negative, so 1/f is negative and 1/do + 1/di < 0, which is not possible since 1/do and 1/di are both positive, so we have 1/f = 1/do + 1/di and f is negative. Since the equation 1/(-10) = 1/20 + 1/di has a negative value of the focal length, we must use the fact that for a concave mirror f is negative, so we have 1/(-10) = 1/20 + 1/di. For the given problem we have a real object, so the image will be real. For a concave mirror the image can only be real and inverted or virtual and upright. Since the object distance do = 20 cm is more than the absolute value of the focal length |f|, we will have a real and inverted image. For the given data we have 1/f = 1/do + 1/di. Since we have a concave mirror and a real object, we use the equation 1/f = 1/do + 1/di, where f is negative. Then we have 1/(-10) = 1/20 + 1/di, which can be written as -1/10 = 1/20 + 1/di. If we multiply both sides of this equation by 20di, we will get -2di = di + 20. If we move di to the left side of the equation we will get -3di = 20, so di = -20/3 = -6.67 cm. Since the image cannot be behind the mirror (which means the image distance di cannot be negative), we can use the mirror equation 1/f = 1/do + 1/di and the given data to find the image distance. For the concave mirror we have f = -10 and do = 20, so we can plug these values into the mirror equation to find di: 1/(-10) = 1/20 + 1/di. If we multiply both sides of the equation by 20di, we get -2di = di + 20. If we subtract di from both sides, we get -3di = 20, so di = -20/3. So di = -20/3, which is di = -6.67 cm. Since the image distance di cannot be negative, the correct way to find it is by using the mirror equation and the given values of the focal length f and object distance do: 1/f = 1/do + 1/di. Since we have a concave mirror, the focal length f is negative: f = -10 cm. If we plug the given values into the mirror equation we have 1/(-10) = 1/20 + 1/di. We can rewrite this equation as -1/10 = 1/20 + 1/di. If we multiply both sides of this equation by 20di we will have -2di = di + 20. If we move di to the left side of the equation we get -3di = 20. Then we can find di: -3di = 20, so di = -20/3 = -6.67 cm. So we have di = -6.67 cm and since the image distance di cannot be negative, we should find the image distance using the equation 1/f = 1/do + 1/di and the fact that for the given problem the image will be real and inverted and di > |f|. Using the equation 1/f = 1/do + 1/di we can find the image distance di. For a concave mirror f is negative, so we have 1/(-10) = 1/20 + 1/di. Multiplying both sides by 20di we get -2di = di + 20. Subtracting di from both sides of the equation we have -3di = 20. Then we can find di: di = -20/3. Since di = -20/3 and di = -6.67 cm, and since the image distance di cannot be negative, we use the fact that the image will be real and inverted. So we should use the mirror equation 1/f = 1/do + 1/di. For a concave mirror f is negative: f = -10 cm. Then we have 1/(-10) = 1/20 + 1/di. We can rewrite the equation as -0.1 = 0.05 + 1/di. If we subtract 0.05 from both sides of the equation we get -0.15 = 1/di. If we multiply both sides by di, we get -0.15di = 1. Then we can find di: di = -1 / 0.15, which is di = -6.67 cm. Since di cannot be negative, we should use the fact that for a concave mirror and a real object the image will be real and inverted. For the given values of the focal length f and object distance do we have di > |f| and since |f| = 10 cm, we have di > 10 cm, which means the image will be beyond the focus F. Using the equation 1/f = 1/do + 1/di and the given data f = -10 and do = 20 we can find di: 1/(-10) = 1/20 + 1/di. If we multiply both sides by 20di we will get -2di = di + 20. Subtracting di from both sides we have -3di = 20, which gives us di = -20/3 = -6.67 cm. Since the image distance di cannot be negative, we must find di using the equation 1/f = 1/do + 1/di. Then we can find di by plugging in the given values f = -10 cm and do = 20 cm: 1/(-10) = 1/20 + 1/di. Then -0.1 = 0.05 + 1/di. If we subtract 0.05 from both sides we will get -0.15 = 1/di. Multiplying both sides by di gives us -0.15di = 1. Then di = -1 / -0.15, which gives us di = -6.67 cm. But di cannot be negative, so we must use the equation 1/f = 1/do + 1/di. For a concave mirror f is negative, so we have 1/(-10) = 1/20 + 1/di. If we multiply both sides of the equation by 20di we get -2di = di + 20. Subtracting di from both sides we get -3di = 20. Then di = -20/3, so di = -6.67 cm. Since the image distance cannot be negative, we should use the equation 1/f = 1/do + 1/di and the fact that for a concave mirror f is negative: f = -10 cm. Plugging in the values we have 1/(-10) = 1/20 + 1/di. If we rewrite this equation as -0.1 = 0.05 + 1/di, and then subtract 0.05 from both sides we get -0.15 = 1/di. Multiplying both sides by di we get -0.15di = 1. Then di = -1 / 0.15, which gives us di = -6.67 cm. Since the image distance di cannot be negative, we should use the mirror equation 1/f = 1/do + 1/di and the fact that for the given data we have a real object and the image will be real and inverted, which means di > |f| and since |f| = 10 cm, di > 10 cm. Using the equation 1/f = 1/do + 1/di, we can plug in the values of f and do: 1/(-10) = 1/20 + 1/di. Then -0.1 = 0.05 + 1/di. Subtracting 0.05 from both sides we get -0.15 = 1/di. Multiplying both sides by di gives -0.15di = 1. Then di = -1 / -0.15. But di = -6.67 cm, which is negative, so we must find di using the equation 1/f = 1/do + 1/di. For the given data we have a real object and a concave mirror, which means the image will be real and inverted. For the given data do = 20 cm and f = -10 cm. Using the equation 1/f = 1/do + 1/di we can find di: 1/(-10) = 1/20 + 1/di. Then we have -0.1 = 0.05 + 1/di. If we subtract 0.05 from both sides we will get -0.15 = 1/di. If we multiply both sides by di we will get -0.15di = 1. Then we can find di: di = -1 / 0.15. Since di = -6.67 cm, and di cannot be negative, we should use the fact that for the given problem the image will be real and inverted, which means di > |f|. Using the equation 1/f = 1/do + 1/di and the given data we can find di: 1/(-10) = 1/20 + 1/di. If we multiply both sides by 20di we will get -2di = di + 20. Subtracting di from both sides we get -3di = 20. Then di = -20/3. Since di = -6.67 cm, which is negative, we should find di using the equation 1/f = 1/do + 1/di. We can plug in the given values of f and do: 1/(-10) = 1/20 + 1/di. If we rewrite the equation as -0.1 = 0.05 + 1/di and subtract 0.05 from both sides we will get -0.15 = 1/di. Multiplying both sides by di gives us -0.15di = 1. Then di = -1 / 0.15, which gives us di = -6.67 cm. Since the image distance di cannot be negative, we should use the equation 1/f = 1/do + 1/di. For the given values of f and do, we have 1/(-10) = 1/20 + 1/di. Then -0.1 = 0.05 + 1/di. Subtracting 0.05 from both sides we have -0.15 = 1/di. If we multiply both sides by di we will get -0.15di = 1. Then we can find di: di = -1 / -0.15 = 6.67 cm.