Refraction and Lenses Legacy Problem #19 Guided Solution

This problem pertains to the formation of an image by a lens. When an image is formed by a lens, there's a relationship between the distance that the object is from the lens, we call that d-object, and the distance the image is formed from the lens, we call that d-image, and the focal length. Here in this question, we're given the d-object. It's 42.8 centimeters. And we're given that the image appears 26.5 centimeters from the lens. That's d-image. In Part A, we're asked to calculate the focal length. The lens equation states that 1 divided by the focal length is equal to 1 divided by d-object plus 1 divided by d-image. So calculating the focal length involves substituting values of d-object and d-image into the lens equation. That becomes 1 over focal length equal 1 over 42.8 centimeters plus 1 over 26.5 centimeters. Evaluating the right side of the equation gives 0.061100 as the value of 1 over focal length. If you take the reciprocal of both sides of the equation, you would have focal length equal 16.3665. This can be rounded to third significant digits, such that focal length equals 16.4 centimeters. In Part B of the problem, we're asked to calculate the image height. The image height is related to the object height, the image distance, and the object distance by the equation high over whole, that's image height over object height, equal negative di over do. Values of di and do are 26.5 centimeters and 42.8 centimeters respectively. And the value of whole, object height, is 4.5 centimeters. So high divided by 4.5 centimeters equal the negative of 26.5 centimeters divided by 42.8 centimeters. Solving this equation for high yields a value of negative 2.7862 centimeters. This can be rounded to two significant digits as negative 2.8 centimeters. The negative of negative 2.8 centimeters simply indicates that the image is an inverted image and as such has a negative image height.