Part A If the arrow is perpendicular to the principal axis of the lens, as in th

Question

Part A If the arrow is perpendicular to the principal axis of the lens, as in the figure (a), what is its lateral imagnification, defined as hi/ho? m= Part B Suppose, instead, that the arrow lies along the principal axis, extending from 72.0cm to 74.1cm from the lens, as indicated in the figure (b). What is the longitudinal magnification of the arrow, defined as Li/Lo? (Hint: Use the thin-lens equation to locate the image of each end of the arrow.) m= Part A If the arrow is perpendicular to the principal axis of the lens, as in the figure (a), what is its lateral imagnification, defined as hi/ho? m= m= m= Part B Suppose, instead, that the arrow lies along the principal axis, extending from 72.0cm to 74.1cm from the lens, as indicated in the figure (b). What is the longitudinal magnification of the arrow, defined as Li/Lo? (Hint: Use the thin-lens equation to locate the image of each end of the arrow.) m= m= m= m=

Explanation / Answer

a) Object distance u =73
Focal length f = -27cm
Let v = image distance
1/v - 1/u = 1/f
Or, 1/v = 1/f + 1/u
Or, 1/v = (f + u)/uf
Or, v = uf/(f + u)------------------(1)
Magnification = v/u = f/(f + u)
= -27/(-27 - 73)
= 0.27
= 0.27

b) For left end:-
Object distance u1 = -74.1 cm
Let v1 = image distance
From (1)
v1 = u1f/(f + u1) = -74.1 * -27/(-27- 74.1) =-19.79cm

For right end:-
Object distance u2 = -72cm
Let v2 = image distance
From (1)
v2 = u2f/(-37 + u1) = -72 * -27/(-27 - 72) =-19.636cm
Length of image = v2 - v = 0.154cm
Length of object = u2 - u1 = 2.1cm
Li/L0 = 0.0733

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---