Physics

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)

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

Explanation- p=100W

_{$\lambda$ 1}= 1nm

_{$\lambda$ 2}= 500nm

Also n₁and n₂ represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n₁hc/ $\lambda$ ₁₌n₂hc/ $\lambda$ ₂

So n₁/ $\lambda$ ₁ = n₂/ $\lambda$ ₂

So n₁/n₂= 1/500

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.