Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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(c) According to VIBGYOR, among all given sources of light, the blue light have the smallest wavelength. According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle. So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

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(b) A passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

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(c) When an object approaches a convergent lens from the left of the lens with a uniform speed of 5 m/s, the image away from the lens with a non-uniform acceleration.

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(d) Since v ∝ λ, the light of red colour is of the highest wavelength and therefore of the highest speed. Therefore, after travelling through the slab, the red colour emerge first.

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(a) since deviation  $\delta =(\mu -1)A$ = (1.5-1)5⁰= 2.5⁰

But $\delta =\theta -r$ so$\theta =\mathrm{7.5\; °}$

This is a short answer type question as classified in NCERT Exemplar

If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis OO'.

Applying lens formula, we have

1/v-1/u=1/f $\frac{1}{\mathit{D}-\mathit{v}}+\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}$ Z

= $\frac{1}{-50}+\frac{1}{25}$

V= 50cm

Magnification m = v/u= 50/-50=-1

So coordinates of image are (50cm, -1cm)

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Since, the object distance is u. Let us consider the two ends of the object be at distance

So u₁= u+L/2 and u₂=u+L/2 respectively so that|u₁- u₂| =L .

 Let the image of the two

ends be formed at v1 and v2, respectively so that the image length would be

L'= |v₁-v₂|……… (1)

Applying mirror formula 1/v+1/u=1/f or v= fu/u-f

On solving two positions of image v₁= $\frac{F(u-L/2)}{u-f-L/2}$ v₂= $\frac{F(u+L/2)}{u-f+L/2}$

For length substituting in equation 1

L'=  |v₁-v₂|= $\frac{F^{2}L}{{(u-F)}^2*\frac{L^2}{4}}$

The objects is short and kept away from focus we have

$\frac{L^2}{4}$ << ( ${u-f)}^2$

Hence L'= $\frac{F^2}{{(u-f)}^2}L$

This is required expression.

This is a short answer type question as classified in NCERT Exemplar

Prism for is given by $\mu =\frac{sin\left\{\frac{A+D_m}{2}\right\}}{sin\left(\frac{A}{2}\right)}$

Given that D_m= A 

After substituting the values, $\mu$ = $\frac{sinA}{sin\frac{A}{2}}$

On solving, $\mu$ = $\frac{2sin\frac{A}{2}cos\frac{A}{2}}{sin\frac{A}{2}}$ 2 cos A/2

So cos A/2 = $\sqrt[]{3}/2$

So A= 60⁰ so this is the required prism angle