Question

JEE mains Previous year question chapter :- Geometrical Optics​

Answer 1

Explanation:

P₁ = -15D and P₂ = +5D

Therefore,,

P(combined) = P₁ + P₂

= -15+5

= -10 D

So, The focal length of the combination is :-

$f = \frac{1}{P}$

$= \frac{1}{ - 10} m$

$= - \frac{1}{10} \times 100 \: cm$

$= - 10 \: cm$

Answer 2

Answer:-

• P1=-15D
• P2=+5D

We know that

$\boxed{\sf P=P_1+P_2}$

$\\ \sf\longmapsto P=-15+5$

$\\ \sf\longmapsto P=-10D$

We know

$\boxed{\sf P=\dfrac{1}{f}}$

$\\ \sf\longmapsto f=\dfrac{1}{P}$

$\\ \sf\longmapsto f=\dfrac{1}{-10}$

$\\ \sf\longmapsto f=-0.1m$

$\\ \sf\longmapsto f=-10cm$

Dora!Dora!