Optical physics question!! I know it's easy, but I'm clueless!?
A photographer uses a lens with f = 60 mm to form an image of a distant object on the CCD detector in a digital camera. The image is 1.0 mm high, and the intensity of light on the detector is 2.1 W/m^2. She then switches to a lens with f = 240 mm that is the same diameter as the first lens. What is the height/intensity of the image now? THANK YOU ANYONE WHO CAN HELP ME!!
The size of the image is proportional to the focal length. So the image would be 240/60=4 times higher and so 4mm
But the intensity is dispersed on a surface which is proportional to 4 squared , and so the intensity by unit surface would be divided by 4^2 =1/16