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Ray Optics and Optical Instruments — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsRay Optics and Optical Instruments3 Marks
Question
What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25\ mm^2$. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
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Answer

Area of the virtual image of each square, $A= 6.25\ mm^2$
Area of each square, $A_0= 1\ mm^2$​​​​​​​
Hence, the linear magnification of the object can be calculated as:
$\text{m}=\sqrt{\frac{\text{A}}{\text{A}_0}}$
$\text{m}=\sqrt{\frac{6.25}{1}}=2.5$
But, $\text{m}=\frac{\text{imagr dis tan ce(v)}}{\text{Object dis tan ce(u)}}$
$\therefore \ \text{v}=\text{mu}$
$= 2.5\ u ...(1)$
Focal length of the magnifying glass, f= 10 cm
According to the lens formula, we have the relation:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{10}=\frac{1}{2.5\text{u}}-\frac{1}{\text{u}}=\frac{1}{\text{u}}\Big(\frac{1}{2.5}-\frac{1}{1}\Big)=\frac{1}{\text{u}}\Big(\frac{1-2.5}{2.5}\Big)$
$\therefore \ \text{u}=\frac{1.5\times10}{2.5}=-6 \ \text{cm}$
and v = 2.5 u
= 2.5 × 6 = -15 cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

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