Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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R
Raj Pandey

Contributor-Level 9

From conservation of Energy:

m g l ( 1 c o s 6 0 ° ) = 1 2 m v 2

v 2 = 2 g l ( 1 2 ) = g l

= 1 0 * 2 5 1 0 = 5 m / s

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Raj Pandey

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| f 1 f 2 | = 4 0 1 2 = 1 0 3

v ( 1 λ 1 1 λ 2 ) = 1 0 3

V = 1 0 3 * λ 1 λ 2 λ 2 λ 1 = 1 0 3 * 4 . 0 8 * 4 . 1 6 . 0 8 = 1 0 3 * 4 0 8 * 4 . 1 6 8 = 7 0 7 . 2 m / s

 

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R
Raj Pandey

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Using factual data.

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R
Raj Pandey

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using factual data.

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Raj Pandey

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Diode conducts current only in one direction

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Raj Pandey

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From Truth table Y = A B ¯ .

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Raj Pandey

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X + p ® Y + b

Q = K Y + K b K P  

Q + K P = K Y + K b  

Q + K P > 0  

 

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Raj Pandey

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λ e = h P e = h 2 m e K e λ e 2 = h 2 2 m e k e K e = h 2 2 m e λ e 2 - (i)

λ P = h P P = h c K P K P = h c λ P - - (ii)

              Ke = KP

λ P α λ e 2

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Raj Pandey

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Resolving power, = d 1 . 2 2 λ = 2 4 . 4 * 1 0 2 1 . 2 2 * 2 4 4 0 * 1 0 1 0  

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R
Raj Pandey

Contributor-Level 9

 At minimum deviation –

δ = 2 i A i = δ + A 2 , r = A 2  

Using snell's low : -         s i n i = μ = s i n r  


δ + A 2 = π A 2 δ = π 2 A  

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