Question

A person is using a concave lens of power -1.5D for distant vision. How has far point of her eye shifted ??

Answer 1

Answer:

Unknown.

Explanation:

P = 1/f

Given : P = - 1.5 D

→ - 1.5 D = 1/f

→ f = 1/(- 1.5)

→ f = - 2/3 or - 0.67 m

Or in other units, focal length

→ - 0.67 × 100 = - 67 cm

Mirror Formula

→ 1/f = 1/v - 1/u

→ 1/(- 67) = 1/v - 1/u

Since the question doesn't mention where the object has placed hence value of v will be in terms of u

→ - 1/67 + 1/u = 1/v

→ (67 - u)/67u = 1/v

→ v = 67u/(67 - u)

Hence far point is surely shifted from ∞.

Answer 2

$\Large \textbf{\underline{\underline{Solution :}}} \\ \\ \sf P = \frac{1}{f} \\ </p><p></p><p> \\ \bf Given : \\ \sf P = - 1.5 D</p><p></p><p> \\ \sf → - 1.5 D = \frac1f</p><p></p><p> \\ \sf → f = \dfrac{1}{(- 1.5)} \\ \sf</p><p></p><p>→ f = - \frac23 \: or \: - 0.67 m \\ \\ \bf Or \: in \: other \: units, \: focal \: length \\ \\ \sf</p><p></p><p>→ - 0.67 × 100 = - 67 cm \\ \\ \large \bf By \: Mirror \: Formula</p><p> \\ \\ \sf</p><p>→ \frac1f = \frac1v - \frac1u</p><p> \\ \sf</p><p>→ \dfrac{1}{(- 67)} = \frac1v - \frac1u \\ \sf</p><p></p><p>→ - \frac{1}{67} + \frac1u = \frac1v \\ \sf</p><p></p><p>→ \frac{(67 - u)}{67u} = \frac{1}{v} \\ \bf</p><p></p><p>→ v = \dfrac{67u}{(67 - u)} \\ \\ \boxed{\textbf {\underline{Hence far point is surely shifted from infinity.}}}$