The graph crosses the y axis at approximately 0.055 this is 1/f 1/f = 0.055 , f = 18.2cm Q33 M = 3, that is v/u = 3 or v = 3u 1/v + 1/u = 1/f 1/3u + 1/u = 1/f 1/3u + 3/3u = 1/f 4/3u = 1/f Now M = 2.5, that is v/u = 2.5 or v = 2.5 u that is (3u -20) = 2.5 x, where x = new object distance. (3u-20)/2.5 = x and 1/v + 1/u = 1/f, Combining and reducing to 1/f = 3.5/(3u-20) substitute for 1/f = 4/3u 10.5 u = 12u - 80 1.5u = 80 u = 53.3 cm therefore 1/f = 4/3u = 4 / 3 x 53.3 f = 40 cm
The graph crosses the y axis at approximately 0.055 this is 1/f 1/f = 0.055 , f = 18.2cm Q33 M = 3, that is v/u = 3 or v = 3u 1/v + 1/u = 1/f 1/3u + 1/u = 1/f 1/3u + 3/3u = 1/f 4/3u = 1/f Now M = 2.5, that is v/u = 2.5 or v = 2.5 u that is (3u -20) = 2.5 x, where x = new object distance. (3u-20)/2.5 = x and 1/v + 1/u = 1/f, Combining and reducing to 1/f = 3.5/(3u-20) substitute for 1/f = 4/3u 10.5 u = 12u - 80 1.5u = 80 u = 53.3 cm therefore 1/f = 4/3u = 4 / 3 x 53.3 f = 40 cm
Solutions-Ch19-RW.docx