For question 11, there is missing information. All the options are possible.

For question 12, no image should be formed.

The second black line is the concave lens.

The red lines, or the rays of light, would diverge upon hitting the concave lens, and the rays would not intersect again. Since in order for another image to be formed after the light hit the concave lens, the 2 rays of light has to intersect, there would be no image formed.

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TopNewestFor question 11, there is missing information. All the options are possible.

For question 12, no image should be formed.

The second black line is the concave lens.

The red lines, or the rays of light, would diverge upon hitting the concave lens, and the rays would not intersect again. Since in order for another image to be formed after the light hit the concave lens, the 2 rays of light has to intersect, there would be no image formed.

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11th Question's answer is 6 cm.

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Please explain .

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Tomorrow.

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@Tanishq Varshney @Chew-Seong Cheong @Kalash Verma @Aditya Kumar @Surya Prakash

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11:-2.25 ?

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I don't know the answer... Anyways how did you reached till there?

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I feel that the 11th question is incomplete.

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This is quite good, as are most Schaum's problem books:

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Thanks sir!

BTW can you help me with question no. 11 & 12?

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11- 6cm 12 - shifts away from the lens system

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Hints for 11 which should be enough to solve it.

Note that linear magnification of a lens is given by $\large \frac{v}{u}$. Linear magnification is length of image divided by length of object.

The second crucial thing to note is that focal length remains unchanged in both the cases. Note that $\large \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

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