Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord!

Like

Report

Numerade Educator

Like

Report

Problem 6 Easy Difficulty

(a) Use conceptual arguments to show that the intensity of light (energy per unit area per unit time) reaching the film in a camera is proportional to the square of the reciprocal of the $f$ -number as
$$I \propto \frac{1}{(f / D)^{2}}$$
(b) The correct exposure time for a camera set to $f / 1.8$ is $(1 / 500)$ s. Calculate the correct exposure time if the $f-$ number is changed to $f / 4$ under the same lighting conditions. Note: "f/4," on a camera, means "an $f$ -number of $4 . "$

Answer

no answer available

Discussion

You must be signed in to discuss.
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Marshall S.

University of Washington

Farnaz M.

Other Schools

Zachary M.

Hope College

Video Transcript

part of the question. The intensity is defined as the amount of energy received per unit area per unit time. So mathematically the relation between intensity and area of them is is given by high equals two. one x 8 one. Okay, so it is proportional intensity is directly proportional to the area. Okay, one. So here is the area of the emmys so let H I. Is horizontal dimension of taking these and h we is the vertical dimension of the image. And assume Q. Is the major distance. And angle made by the light ray with the principal axis. So we can write the angle. Theta equals two. Actually they were by Q as area angle is not changing. So we can write shh edge, horizontal dimension is directly proportional to cuba. Hence the area of the image is proportional to the square of the distance. Now if the object is very far from the lens then the major distance is equal to the focal length. Okay, so now we can write that area of the image will be directly proportional to atmosphere. Okay, so where is the focal length of the lens? So now combining this relation? This relation and this relation. We can write intensity I is directly proportional to one by atmosphere. Okay, so now also the intensity of the light received is directly proportional to the square of diameter. So we can write I directly proportional to the square. Okay, so now combining these equations one and two, we get intensity I is directly proportional to the square by atmosphere. And it can be written as I directly proportional to one by F by the police fear. Okay, so this is the relation which we want to prove in the party of the problem. Okay, so this is the answer for the party of the problem. Now moving to the party in which we want to we want the product of citrus peel and area of the lens to be constant. Show the area of the lens is proportional to the square, so it gives shattering spear as one divided by saturday spear as two equals two. F. Number one divided by F. Number two. Okay, it's clear. So no uh shutter speed to as two will be equals two. F one, F two, F one and F two R. F. Number for the 1st and 2nd lines square multiplied by a satirist essence, substituting the values we get as two equals to four by 1.8 square, multiplied by one by 500. These values are given in this question from here, Saturday speed to comes out before dinner before 05 second. Okay, so this is the answer for the question.

Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Marshall S.

University of Washington

Farnaz M.

Other Schools

Zachary M.

Hope College