 # A Point Object Is Placed On The Principal Axis Of The Convex Lens Of Focal Length 20Cm At A Distance 40 Cm To The Left On It. The Diameter Of The Lens Is 10Cm. If The Eye Is Placed 60Cm To The Right Of The Lens At A Distance H Below The Principal Axis, Then Max Value Of 'h' To See The Image Is____.

Solution:

Given:

Distance of object, u= − 40cm

focal length, f = 20cm

Now by applying the lens formula, we get

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \\ \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \\ \Rightarrow \frac{1}{v} = \frac{1}{20} + \frac{1}{(-40)}\\\Rightarrow \frac{1}{v} = \frac{2-1}{40} \\\Rightarrow V = 40cm$$

Therefore, the image is formed at the curvature.

As given in the question:

$$\Delta CDE \; and \; \Delta CAB$$ are symmetric

So, $$\frac{AB}{DB} = \frac{BC}{EC}$$ $$\Rightarrow \frac{h}{5} = \frac{20}{40} \\\Rightarrow h = \frac{5}{2} \\\Rightarrow h = 2.5cm$$

Therefore, the maximum value of h to see the image is 2.5cm.

Explore more such questions and answers at BYJU’S. (0) (0)