MCQ
prism can produce a minimum deviation $\delta$ in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is:
- A$0$
- ✓$\delta$
- C$2\delta$
- D$3\delta$
In combination $($refractive angles of prisms reversed with respect to each other$)$, the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.
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$(i)\;A+B\;\to\; C \;+\;\varepsilon $
$(ii)\;C\;\to \;A\;+\;B\;+\;\varepsilon $
$(iii)\;D\;+\;E\;\to \; F\;+$$\;\varepsilon $
$(iv)\;F\;\to \; D\;+\;E\;+\;\varepsilon $
$\;\varepsilon $ is the energy released? In which reactions is $\;\varepsilon $ positive?
$(i)$ Electrons $(ii)$ Protons $(iii)$ $H{e^{2 + }}$ $(iv)$ Neutrons
The emission at the instant can be
