As shown in the figure- a parallel beam emerges parallel- The relation between mu - mu - 1 -and - mu - 2 - is- dfrac - 1 - mu - -dfrac - 1 - - mu - 1 - - -dfrac - 1 - - mu - 2 - - mu - mu - 1 - - mu - 2 - dfrac - 2 - mu - -dfrac - 1 - - mu - 1 - - -dfrac - 1 - - mu - 2 - - mu -dfrac - - mu - 1 - mu - 2 - - 2 -

Question

As shown in the figure- a parallel beam emerges parallel- The relation between  mu - mu - 1 -and  - mu - 2 - is- dfrac - 1 - mu - -dfrac - 1 - - mu - 1 - - -dfrac - 1 - - mu - 2 - -  mu - mu - 1 - - mu - 2 - dfrac - 2 - mu - -dfrac - 1 - - mu - 1 - - -dfrac - 1 - - mu - 2 - -  mu -dfrac - - mu - 1 - mu - 2 - - 2 -

Toppr Admin · Accepted Answer

We have t-x221D-x3BC-

As shown in the figure, a parallel beam emerges parallel. The relation between $μ, μ_{1}$ and $μ_{2}$ is:

$\frac{1}{μ} = \frac{1}{μ_{1}} + \frac{1}{μ_{2}}$
$μ = μ_{1}$ + $μ_{2}$
$\frac{2}{μ} = \frac{1}{μ_{1}} + \frac{1}{μ_{2}}$
$μ = \frac{μ_{1} + μ_{2}}{2}$

We have
$t \propto μ$

As shown in the figure, a parallel beam emerges parallel. The relation between μ,μ1and μ2 is:1μ=1μ1+1μ2μ=μ1+ μ22μ=1μ1+1μ2μ=μ1+μ22

We have t∝μ

As shown in the figure, a parallel beam emerges parallel. The relation between $μ, μ_{1}$ and $μ_{2}$ is:

$\frac{1}{μ} = \frac{1}{μ_{1}} + \frac{1}{μ_{2}}$
$μ = μ_{1}$ + $μ_{2}$
$\frac{2}{μ} = \frac{1}{μ_{1}} + \frac{1}{μ_{2}}$
$μ = \frac{μ_{1} + μ_{2}}{2}$

We have
$t \propto μ$