Physics 2 Marks Questions

u = -20cm, v = -50cm

from lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{f}}=\frac{1}{-50}-\frac{1}{-20}=\frac{1}{-20}-\frac{1}{50}=\frac{3}{100}$

$\Rightarrow\text{f}=\frac{100}{3}\text{cm}=\frac{1}{3}\text{m}$

So, power of the lens $=\frac{1}{\text{f}}=3\text{ Diopter}$

$\text{u}= \infty$ and $\text{v}=-200\text{cm}=-2\text{m}$

So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-2}-\frac{1}{\infty}$

$\frac{1}{\text{f}}=-\frac{1}{2}=-0.5$

So, power of the lens is -0.5D

F_e = 10cm, L = 1m = 100cm

S0, f₀ = L - f_e = 100 - 10 = 90cm

and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$

Because, the eye is relaxed the image is formed at infinity (normal adjustment)

So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$

For the relaxed farsighted eye, $\text{D} = 40\text{cm}$

So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$

So, its magnifying power is 8X.

So, the person must be near sighted

$\text{u}=\infty,$ $\text{v}=\text{far point,}$ $\text{f}=\frac{1}{-2.5}=-0.4\text{m}=-40\text{cm}$

Now, $\frac{1}{\text{v}}-\frac{1}{\text{v}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=0+\frac{1}{-40}$

$\Rightarrow \text{v}=-40\text{cm}$

So, the far point of the person is 40cm.