Question

Single Correct Answer Type
QUESTION 41:
A bright object and a viewing screen are separated by a distance of 76.0 cm. At what location(s) between
the object and the screen should a lens of focal length 16.0 cm be placed in order to produce a crisp
image on the screen?
0 42.9 cm
© 22.9 cm
o 32.9 cm
o 52.9 cm​

Answer 1

Given: Distance between object and viewing screen (u + v )= 76cm

            Focal length of lens f  =  16cm

To find:  

Location of lens between object and the screen

            u + v = 76 cm

            u = 76 - v -----------(1)

From the lens formula-

⇒   1/u + 1/v = 1/f

⇒  v + u/ uv = 1/f

Subtituting the value of u from ....(1)

(v + 76 - v) / (76 - v)v = 1/f

76 / (76v - v²) = 1 / 16

1216 = 76v - v²

v² - 76v + 1216 = 0

By solving the quadratic equation

v = 22.9 cm

Therefore, the lens should be placed at a distance of 22.9cm between the object and the screen

Answer 2

Answer:

22.9 cm

Explanation:

The sum of the object and image d_i stances must be the distance

between object and screen

$d_{0}+d_{i}=L=76.0 \mathrm{cm}$

Focal length of the lens

From the lens makers formula

$\begin{array}{c}\left(\frac{1}{\mathrm{d}_{0}}\right)+\left(\frac{1}{d_{i}}\right)=\frac{1}{f} \\\frac{1}{d_{0}}+\left[\frac{1}{\left(76.0 \mathrm{cm}-d_{0}\right)}\right]=\frac{1}{16.0 \mathrm{cm}} \\d_{0}^{2}-d_{0} 76.0 \mathrm{cm}+1216 \mathrm{cm} =0\\d_{0}=54.6 \mathrm{cm}, 22.9 \mathrm{cm}\end{array}$

smallest possible distance is $22.9 cm$