
$\text{R}_1+\text{ma}-\text{mg}=0$
$\Rightarrow\text{R}_1=\text{m(g}-\text{a) = mg}-\text{ma} \ ...(\text{i})$
$\text{T}-\mu\text{R}_1=0\Rightarrow\text{T = m(mg}-\text{ma}) \ ...(\text{ii})$
Again, $\text{F}-\text{T}-\mu\text{R}_1=0$
$\Rightarrow\text{F}-\{\mu(\text{mg}-\text{ma})\}-\text{u}(\text{mg}-\text{ma})=0$
$\Rightarrow\text{F}-\mu\text{mg}+\mu\text{ma}-\mu\text{mg}+\mu\text{ma}=0$
$\Rightarrow\text{F}=2\mu\text{mg}-2\mu\text{ma}\Rightarrow\text{F}=2\mu\text{m(g}-\text{a})$

$\text{R}_1=\text{mg}-\text{ma} \ ...(\text{i})$
$2\text{F}-\text{T}-\mu\text{R}_1-\text{ma}_1=0$
$\Rightarrow2\text{F}-\text{t}-\mu\text{mg}+\mu\text{a}-\text{ma}_1=0 \ ...(\text{ii})$

$\text{T}-\mu\text{R}_1-\text{Ma}_1=0$
$\Rightarrow\text{T}=\mu\text{R}_1+\text{Ma}_1$
$\Rightarrow\text{T}=\mu(\text{mg}-\text{ma})+\text{Ma}_1$
$\Rightarrow\text{T}=\mu\text{mg}-\mu\text{ma + Ma}_1$
Subtracting values of F & T, we get
$2(2\mu\text{m(g}-\text{a}))-2(\mu\text{mg}-\mu\text{ma}+\text{Ma}_1)\\-\mu\text{mg}+\mu\text{ma}-\mu\text{a}_1=0$
$\Rightarrow4\mu\text{mg}-4\mu\text{ma}-2\mu\text{mg}+2\mu\text{ma = ma}_1+\text{Ma}_1$
$\Rightarrow\text{a}_1=\frac{2\mu\text{m(g}-\text{a})}{\text{M + m}}$
Both blocks move with this acceleration but in opposite directions.
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Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I0 is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.


