Question

Playing with a lens one morning, Rita discovers that if she holds the lens 0.120 m away from a wall opposite to a window, she can see a sharp but upside-down picture of the outside world on the wall. That evening, she covers a lighted lamp with a piece of card on which she has pierced a small hole, 0.005 m in diameter. By placing the lens between the illuminated card and the wall, she manages to produce a sharp image of diameter 0.020 m on the wall. What is the distance between the card and the wall?

• A. 0.450 m
• B. 0.750 m
• C. 0.600 m
• D. 0.300 m

Answer 1

Answer: (B)

0.750 m

We assume the sign convention where distance measured in the direction of light
propagation are positive, and in opposite direction, negative. For the morning experiment, since the image of the outside world $(objectdistanceu\approx inf)$ is real, sharp and inverted, the lens must be converging one, and the image distance $\left(v\right)$ is equal to the focal length $\left(f\right)$ of the lens.Thus $f=0.120m$. for the evening experiment, the hole in the card serves as the object of size 0.005m. The image size on the wall is 0.020m.

• Distance between the lens and the card $u=-uc(uc>0)$
• Distance between the lens and the wall $v=+vc(vc>0)$
• Distance between the lens and wall (question) $d=uc+vc$
• $Magnification=vc/-uc=0.020/-0.005=-4$(real images by converging lenses always have negative magnification, i., inverted image).

Thus, applying the lens formula
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{vc}-\frac{1}{uc}=\frac{1}{f}$
$\frac{1}{4uc}+\frac{1}{uc}=\frac{1}{f}$
$\frac{5}{4uc}=\frac{1}{f}$

$uc=5f/4=5\times 0.120m/4=0.150m$
$vc=4uc4\times 0.150m=0.600m$
Therefore, the distance between the lens and the wall is
$d=uc+vc=(0.150+0.600)m=0.750m$