Q 9.25  What should be the distance between the object in figure 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

                

Answers (1)

Given

Virtual image area = 6.25 mm2

Actual ara = 1 mm2

We can calculate linear magnification as 

m=\sqrt{\frac{6.25}{1}}=2.5

we also know 

m=\frac{v}{u}

v = mu

Now, according to the lens formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

\frac{1}{10}=\frac{1}{mu}-\frac{1}{u}

\frac{1}{u}(\frac{1}{2.5}-1)=\frac{1}{10}

u=-6cm and 

v=mu=2.5*(-6)=-15cm

Since the image is forming at a distance which is less than 25 cm, it can not be seen by eye distinctively.

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