A prism - mu - 1-5- has the refracting angle of 60-o- The deviation of a monochromatic ray incident normally on its one surface will be -sin 48-o 36- - 0-75-36-o 72-20-o 30-18-o22-o 1-

Question

Toppr Admin · Accepted Answer

For a prism-x3BC-sin-A-x3B4-2-sin-A2-1-5-xD7-0-5-sin-A-x3B4-2-A-x3B4-2-48o36-x2032-x3B4-36o72-x2032-

A prism $(μ = 1.5)$ has the refracting angle of $60^{o}$ . The deviation of a monochromatic ray incident normally on its one surface will be $(s i n 48^{o} 36^{'} = 0.75)$
$36^{o} 72^{'}$
$20^{o} 30^{'}$
$18^{o}$
$22^{o} 1^{'}$

For a prism,
$μ = \frac{sin (\frac{A + δ}{2})}{sin (\frac{A}{2})}$
$1.5 \times 0.5 = sin (\frac{A + δ}{2})$
$\frac{A + δ}{2} = 48^{o} 36^{'}$
$δ = 36^{o} 72^{'}$

A prism (μ=1.5) has the refracting angle of 60o. The deviation of a monochromatic ray incident normally on its one surface will be (sin48o36′=0.75)36o72′20o30′18o22o1′

For a prism,μ=sin(A+δ2)sin(A2)1.5×0.5=sin(A+δ2)A+δ2=48o36′δ=36o72′

A prism $(μ = 1.5)$ has the refracting angle of $60^{o}$ . The deviation of a monochromatic ray incident normally on its one surface will be $(s i n 48^{o} 36^{'} = 0.75)$
$36^{o} 72^{'}$
$20^{o} 30^{'}$
$18^{o}$
$22^{o} 1^{'}$

For a prism,
$μ = \frac{sin (\frac{A + δ}{2})}{sin (\frac{A}{2})}$
$1.5 \times 0.5 = sin (\frac{A + δ}{2})$
$\frac{A + δ}{2} = 48^{o} 36^{'}$
$δ = 36^{o} 72^{'}$