Question

the angle of prism of refractive index 1.5 is 60 degree. what will be the angle of minimum deviation of prism. sin 49 =0.75​

Answer 1

$\bigstar$ $\sf{\red{\underline{\underline{\purple{Correct\:Answer\::38°}}}}}$

$\bigstar$ $\sf{\green{\underline{\underline{\orange{CONCEPTS:-}}}}}$

☞ If “Angle of Prism (A)” , “Refractive index (n)” is given and asked to find “Angle of minimum deviation ” , then formula will be

• $\tt{n\:=\:{\dfrac{\sin({\dfrac{A\:+\:{\delta_{min}}}{2}})}{\sin({\dfrac{A}{2}})}}}$

Where

1. n = refractive index
2. A = angle of Prism
3. ${\delta_{min}}$ = angle of minimum deviation

$\bigstar$ $\sf{\green{\underline{\underline{\orange{To\:Find:-}}}}}$

☞ Angle of minimum deviation of prism

$\bigstar$ $\sf{\green{\underline{\underline{\orange{SOLUTION:-}}}}}$

☞ GIVEN :-

• n = 1.5
• A = 60
• sin (49°) = 0.75

Now we have know that,

$\tt{n\:=\:{\dfrac{\sin({\dfrac{A\:+\:{\delta_{min}}}{2}})}{\sin({\dfrac{A}{2}})}}}$

$\tt{\implies\:1.5\:=\:{\dfrac{\sin({\dfrac{60\:+\:{\delta_{min}}}{2}})}{\sin({\dfrac{60}{2}})}}}$

$\tt{\implies\:1.5\:\times\:0.5\:=\:{\sin({\dfrac{60\:+\:{\delta_{min}}}{2}})}}$

$\tt{\implies\:{\sin(49)}\:=\:{\sin({\dfrac{60\:+\:{\delta_{min}}}{2}})}}$

$\tt{\implies\:49\:=\:{\dfrac{60\:+\:{\delta_{min}}}{2}}}$

$\tt{\implies\:98\:=\:60\:+\:{\delta_{min}}}$

$\tt{\implies\:{\delta_{min}}\:=\:98\:-\:60}$

$\tt{\implies\:{\delta_{min}}\:=38°}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::38°}}}}$

Answer 2

Answer:

n= 1.5

A= 60°

sin 49°

• n= sin (A+Smin )
• 2.
• sin A
• 2

1.5= sin (60+Smin)

2.

sin 60

2

1.5×0.5 sin(60+Smin)

sin (49°) = sin (60+Smin)

2

49 = 60+Smin

2

98= 60+Smin

Smin = 98-60

Smin = 38°

THE ANSWER IS 38°