Question

The thin lenses of focal length 25cm and -50cm are placed in contact with each other . Explain wether the combination will behave like a converging or diverging lens . Find power and focal length of the combination​

Answer 1

★Given :-

• f 1 = 25 cm
• f 2 = -50 cm

★To find :-

• Focal length of combination lens
• Nature of combination  lens
• Power of combination lens

★Solution :-

First ,let's find the focal length of the combination lens

$\implies\mathtt{\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}}$

$\implies\mathtt{\dfrac{1}{f}=\dfrac{1}{25}+\dfrac{1}{-50}}$

$\implies\mathtt{\dfrac{1}{f}=\dfrac{1}{25}-\dfrac{1}{50}}$

$\implies\mathtt{\dfrac{1}{f}=\dfrac{2-1}{50}}$

$\implies\mathtt{\dfrac{1}{f}=\dfrac{1}{50}}$

$\implies\boxed{\mathtt{f = 50 cm }}$

As the focal length of the combination lens is positive this means that the combination lens will behave as a convex lens and it will be converging in nature

Power of the combination lens is given by the formula ,

$\implies\mathtt{P=\dfrac{1}{f(m)}}$

First we need to convert the focal length into m in order to to that simply divide by 100

$\implies\mathtt{f = \dfrac{100}{50} m}$

Hence,

$\implies\mathtt{P=\dfrac{100}{50}}$

$\implies\boxed{\mathtt{P=+ 2D }}$

★Points to remember :-

• Image distance is always negative as distance is measured from the pole of the mirror/ the optical Centre of the lens
• Focal length of a concave mirror = - ve
• Focal length of convex mirror = + ve
• Focal length of a concave lens = - ve
• Focal length of convex lens = + ve

★Formulas :-

♣Mirror :-

• Mirror formula ,

$\implies\boxed{\mathtt{\dfrac{1}{f} =\dfrac{1}{v} +\dfrac{1}{u} }}$

• Magnification

$\implies\boxed{\mathtt{m=-\dfrac{v}{u} =\dfrac{h_i}{h_o} }}$

♣Lens:-

• Lens formula ,

$\implies\boxed{\mathtt{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }}$

• Magnification

$\implies\boxed{\mathtt{m=\dfrac{v}{u} =\dfrac{h_i}{h_o} }}$