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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Ray Optics

BITSAT 2022 · Physics · Ray Optics

An achromatic convergent doublet of two lense in contact has a...

An achromatic convergent doublet of two lense in contact has a power of +5D+5 \mathrm{D}. The power of converging lens is +6D+6 \mathrm{D}. The ratio of the dispersive power of the convergent and divergent lenses is

Options

  1. A3:73: 7
  2. B2:32: 3
  3. C1:51: 5
  4. D5:35: 3

Correct answer

C. 1:51: 5

Step-by-step solution

- The condition of achromatism is W1P1+W2P2=0W_1 P_1+W_2 P_2=0⇒W1P1=−W2P2⇒W1W2=−P2P1\begin{array}{rlrl} \Rightarrow& W_1 P_1 & =-W_2 P_2 \\ \Rightarrow & \frac{W_1}{W_2}& =-\frac{P_2}{P_1} \end{array}Now,Now,\quad P_1+P_2=4 D\mathrm{D} but, Power of converging lens, P_1=5 D\mathrm{D}\becausePowerofdiverginglensPower of diverging lensP2=4D−P1=4D−5D=−D\begin{aligned} P_2 & =4 \mathrm{D}-P_1 \\ & =4 \mathrm{D}-5 \mathrm{D}=-\mathrm{D} \end{aligned}[Fromii]∴[From ii] \therefore From Eq. (i), we have W1W2=−P2P1=−(−D)5D=15\frac{W_1}{W_2}=-\frac{P_2}{P_1}=\frac{-(-D)}{5 D}=\frac{1}{5} \Rightarrow $\frac{W_1}{W_2}=\frac{1}{5}
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