An object is placed at 5.6 x 10-4m in front of a converging lens of focal length 1.0 x 10-nm. The image formed is

Answer Details

The given problem involves a converging lens, which means it has a positive focal length. The object is placed at a distance of 5.6 x 10^-4 m from the lens. Using the lens formula (1/f = 1/v - 1/u), we can find the image distance (v). 1/f = 1/v - 1/u 1/1.0 x 10^-n = 1/v - 1/5.6 x 10^-4 On simplifying the above equation, we get: v = (1.0 x 10^-n)(5.6 x 10^-4) / (1.0 x 10^-n + 5.6 x 10^-4) We can see that the image distance (v) depends on the value of n. However, regardless of the value of n, we can still determine the nature of the image formed by considering the sign of v. If v is positive, the image is real and located on the opposite side of the lens as the object. If v is negative, the image is virtual and located on the same side of the lens as the object. In this case, the object is placed on the left side of the lens and the lens is converging. Therefore, the image formed will be real and located on the right side of the lens. So, the options (2) and (4) can be eliminated. Now, we need to determine the orientation and size of the image. For this, we can use the magnification formula (m = -v/u), where u is the object distance. m = -v/u = -[(1.0 x 10^-n)(5.6 x 10^-4) / (1.0 x 10^-n + 5.6 x 10^-4)] / (5.6 x 10^-4) On simplifying the above equation, we get: m = -1 / (1 + 1.0 x 10^n/5.6) We can see that the magnification (m) is negative, which means that the image is inverted with respect to the object. Also, the magnitude of the magnification depends on the value of n. However, regardless of the value of n, we can still determine the relative size of the image by considering its absolute value. If |m| > 1, the image is magnified. If |m| < 1, the image is diminished. If |m| = 1, the image is the same size as the object. In this case, the magnitude of the magnification is greater than 1 (i.e., |m| > 1), which means that the image is magnified. Therefore, the correct option is (3) Real, inverted and magnified.