The refracting angle of the prism is - 4-5 - circ - and its refractive index is 1-52- The angle of minimum deviation will be - - 2 - circ - - 2-3 - circ - - 1-5 - circ - - 4-5 - circ -

Question

The refracting angle of the prism is  - 4-5 - circ - and its refractive index is 1-52- The angle of minimum deviation will be - - 2 - circ - - 2-3 - circ - - 1-5 - circ - - 4-5 - circ -

Toppr Admin · Accepted Answer

Given-A-4-5-x2218-x3BC-1-52The minimum deviation is given by -x3B4-min-x3BC-x2212-1-xD7-A-gt- -x3B4-min-1-52-x2212-1-xD7-4-5-2-34-x2218-

The refracting angle of the prism is ${4.5}^{\circ}$ and its refractive index is $1.52$ . The angle of minimum deviation will be :
$2^{\circ}$
${2.3}^{\circ}$
${1.5}^{\circ}$
${4.5}^{\circ}$

Given,
$A = {4.5}^{\circ}$

$μ = 1.52$

The minimum deviation is given by :
$δ_{m i n} = (μ - 1) \times A$

=> $δ_{m i n} = (1.52 - 1) \times 4.5 = {2.34}^{\circ}$ .

The refracting angle of the prism is 4.5∘ and its refractive index is 1.52. The angle of minimum deviation will be :2∘2.3∘1.5∘4.5∘

Given,A=4.5∘μ=1.52The minimum deviation is given by :δmin=(μ−1)×A=> δmin=(1.52−1)×4.5=2.34∘.

The refracting angle of the prism is ${4.5}^{\circ}$ and its refractive index is $1.52$ . The angle of minimum deviation will be :
$2^{\circ}$
${2.3}^{\circ}$
${1.5}^{\circ}$
${4.5}^{\circ}$

Given,
$A = {4.5}^{\circ}$

$μ = 1.52$

The minimum deviation is given by :
$δ_{m i n} = (μ - 1) \times A$

=> $δ_{m i n} = (1.52 - 1) \times 4.5 = {2.34}^{\circ}$ .