A student set up the magnetic optical bench as shown. The locations of the compo
ID: 1551745 • Letter: A
Question
A student set up the magnetic optical bench as shown. The locations of the component holders for the crossed arrow target (CAT), and slens (L), are as follows, respectively: 228 mm, 579 mm. The offsets for the CAT and the S are 4.00 mm, and the offset for lens, L is 6.00 mm. The inverted image formed on the screen is 0.750 times the size of the object. (a) What is the position of the CAT on the optical bench? (b) What is the position of the L on the optical bench? (c) Calculate the position of the component holder of the screen? (d) What is the position of the s on the optical bench? (e) Calculate the object distance from the lens (f Calculate the image distance from the lens. (g) Calculate the focal length of the lens. (d) Without changing the CAT and the S location, calculate position where would you place the component holder of the lens to obtain second image ? Please show detailed work for all answers.Explanation / Answer
Accordin to given condition,
Object distance xo = 228 mm
image distance xi = 579 mm
let, position of compound holder of screen = x
object distance from lens do = x - xo
image distance from lens di = xi - x
(a)
position of CAT = 228 + 4 = 232 mm
(c)
We know that,
Magnification M = di / do
0.75 = (xi - x) / (x - (-xo))
0.75x + 0.75xo = xi - x
by solving,
x = 233.14 mm
(d)
position of S = 579 + 6 = 585 mm
(e)
object distance from lens do = x -xo = 233.14 - 228 = 5.14 mm
(f)
image distance from lens di = xi - x = 579 - 233.14 = 345.86 mm
(g)
Let, focal length of lens = f
From the lens formula,
1 / f = 1 / di - 1 / do
1 / f = 1 / 345.86 - 1 / (-5.14)
by solving,
f = 5.064 mm
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