Question

what is the focused distance of a combined lens made by combining a concave lens of 30cm focused distance placed in contact with a convex lens of 20cm focus distance ? wheather this combination is convergent lens or divergent lens .​

Answer 1

SOLUTION

GIVEN

A concave lens of 30 cm focus distance placed in contact with a convex lens of 20 cm focus distance

TO DETERMINE

• The focus distance of the combined lens

• Whether this combination is convergent lens or divergent

EVALUATION

Here it is given that a concave lens of 30 cm focus distance placed in contact with a convex lens of 20 cm focus distance

We know that Focus distance of a convex lens is positive and of concave lens is negative

$\sf{f_1 = Focus \: distance \: of \: concave \: lens = - 30}$

$\sf{f_2= Focus \: distance \: of \: convex \: lens = + 20}$

Let f = The focus distance of the combined lens

We are aware of the formula that

$\displaystyle \sf{ \frac{1}{f} = \frac{1}{f_1 } + \frac{1}{f_2 } }$

$\displaystyle \sf{ \implies \: \frac{1}{f} = \frac{1}{ - 30 } + \frac{1}{20 } }$

$\displaystyle \sf{ \implies \: \frac{1}{f} = - \frac{1}{ 30 } + \frac{1}{20 } }$

$\displaystyle \sf{ \implies \: \frac{1}{f} = \frac{ - 2 + 3}{60} }$

$\displaystyle \sf{ \implies \: \frac{1}{f} = \frac{ 1}{60} }$

$\displaystyle \sf{ \implies \: f = 60 }$

The focus distance of the combined lens = 60 cm

Again

Power of the concave lens

$\displaystyle \sf{P_1 = - \frac{1}{0.30} = - \frac{10}{3} }$

Power of the convex lens

$\displaystyle \sf{P_2 = \frac{1}{0.20} = 5 }$

So the power of the combined lens P is given by

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

$\displaystyle \sf{ \implies \: P = - \frac{10}{3} + 5 }$

$\displaystyle \sf{ \implies \: P = \frac{ - 10 + 15}{3} }$

$\displaystyle \sf{ \implies \: P = + \frac{ 5}{3} }$

Since focus distance and power of the combined lens are both positive

So combined lens is convergent

FINAL ANSWER

• The focus distance of the combined lens = 60 cm

• The combined lens is convergent

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