one converging lens and diverging lens of focal lengths $30 cm$ and $10 cm r e s p e c t i v e l y$ , are placed at some seperation. A parallel beam of light is incident of $1^{st}$ lens then

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For emergent beam from $2^{n d}$ lens to be parallel the seperation between the lenses has to be $40 c m$

If lenses are placed at such a seperation that emergent beam from $2^{n d}$ lens is parallel, then emergent beam width is $4 c m$ if original beam width is $1 c m$ .

If lenses are placed at such a seperation that emergent beam from $2^{n d}$ lens is parallel. then emergent beam width is $1 c m$ if original beam has a width of $3 c m$ .

for emergent beam from $2^{n d}$ lens to be parallel the seperation between the lenses has to be $20 c m .$

answer is A, C.

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Detailed Solution

In this case , the second focal point of the convex lens must be the first focal point of the concave lens .

Therefore , separation d = $f_{1} - f_{2} = (30 - 10) c m = 20 c m$

If R be the radius of the incident beam and r be the radius of the emerent beam , then $\frac{r}{10} = \frac{R}{30} \Rightarrow r = \frac{10}{30} \times 1.5 c m = 0.5 c m .$