# A prism can produce a minimum deviation d in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is: Option 1) 0 Option 2) d Option 3) 2d Option 4) 3d

Condition of maximum deviation -

$i=e$

- wherein

$\delta _{min}= 2i-A$

$\mu = \frac{\sin \left ( \frac{\delta _{m}+A}{2} \right )}{\sin \left ( A/2 \right )}$

value of a minimum deviation

$d_{net} = d-d+d =d$

Option 1)

0

Incorrect

Option 2)

d

Correct

Option 3)

2d

Incorrect

Option 4)

3d

Incorrect

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