Physics

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

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 Questions as classified in NCERT Exemplar

Explanation- In photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.Thus, all the electrons that absorb a photon doesn't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.

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

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

apparent depth = real depth / refractive index

Since, the image formed by Medium1, O₂ act as an object for Medium2.

If seen from µ₃ ,the apparent depth is O₂.

Similarly, the image formed by Medium2, O₂ act as an object for Medium3

O₂₌$\frac{{\mu }_3}{{\mu }_2}$ ( $\frac{h}{3}+O_1$ )

= $\frac{{\mu }_3}{{\mu }_2}$ ( $\frac{h}{3}+\frac{{\mu }_2}{{\mu }_1}\frac{h}{3})$

Seen from outside the apparent height is

O₃ = $\frac{1}{{\mu }_3}$ (  $\frac{h}{3}+O_2)$ = $\frac{1}{{\mu }_3}\left(\frac{h}{3}+\frac{h}{3}\left(\frac{{\mu }_3}{{\mu }_2}+\frac{{\mu }_3}{{\mu }_1}\right)\right)$

= $\frac{h}{3}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3}$ )

This is the expression for required depth. 

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

no reversibility of lens maker formula is not possible.

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

m=v/u = D/f

So m = 25/10 = 2.5= 0.025m

P= 1/0.025= 40D

This is the required power of lens.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- (i) Here it is given that, an electron absorbs 2 photons each of frequency ν then ν where, v′ is the frequency of emitted electron.

Given, E_max= hv- ${?}_0$

Now, maximum energy for emitted electrons is E_max= h2v- ${?}_0=2\mathrm{h}\mathrm{v}-{?}_0$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible

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

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

By lens makers formula 1/f= $(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

So f_br so focal length of blue is less than red light 

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- $\lambda =h/\sqrt[]{2mqv}$

$\lambda \propto \frac{1}{\sqrt[]{mq}}$

$\frac{{\lambda }_p}{{\lambda }_a}=\frac{\sqrt[]{m_a{q}_a}}{\sqrt[]{m_p{q}_p}}=\frac{\sqrt[]{4m_p*2e}}{\sqrt[]{m_p*e}}$ = $\sqrt[]{8}$

So wavelength of proton is $\sqrt[]{8}$ times of alpha particle