
#1
Sep2705, 11:28 AM

P: 4

Could somebody please help me on the following questions:
1. A camera with a 50mm focal length lens focuses on a man 1.8m tall standing 40m from the camera. (a) What angle does the man subtend at the camera lens? [I do not know what subtend means] (b) What is the size of the image of the man on the film? 



#2
Sep2705, 11:46 AM

HW Helper
P: 930

Try a Google search on 'subtend'.
hotvette 



#3
Sep2805, 04:31 PM

HW Helper
P: 1,449

They want to know what angle the top of the man makes at the centre of the lens. This means that
[tex]\tan\theta=\frac{1.8}{40}[/tex] since the base is 40m and the perpendicular is 1.8m of the right angle that the man subtends at the centre of the lens. To calculate the image size you need to first calculate the image distance with the lens formula and then use the fact that the image will subtend the same angle [tex]\theta[/tex] at the centre of the lens since the ray from the top of the man continues straight through the centre of the lens to the film. With this image distance and the known angle you can then again calculate the image size. Actually since the object is so far away from the lens we can simplify the problem  first lets look at the right handed triangle on the object side of the lens: [tex]\tan \theta=\frac{1.8}{40}=\frac{18}{400}=\frac{9}{200}[/tex] Furthermore, since the object is relatively far away we know that the image will form in the focal plane of the lens. On the image side of the lens we can therefore say that: [tex]\tan \theta = \frac{h}{50} \therefore h=\frac{50\times9}{200}=\frac{9}{4}=2.25mm[/tex] 


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