Physics

CBSE Class 12

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41.

When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why

A bright spot is seen at the centre of the shadow of objective because wave diffracted from the edge of the circular obstacle interface constructively at the centre of the shadow producing a bright spot.

42.

Obtain the mirror formula and write the expression for the linear magnification.

A relationship among the object distance (u), the image distance (v) and the focal length (f) of a mirror are called the mirror formula.

The formula is given by $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$

Take an object AB beyond C of a concave mirror MM'. A ray AD parallel to principal axis passes through focus after reflection.

Another ray AE which is passing through C comes back along the same path after reflection.

These two reflected rays intersect at A'. A' draw perpendicular A'B' on the principal axis. So A'B' is a real and inverted image which is formed between C and F which is smaller than the object in size.

Draw DG perpendicular to the principal axis. So, applying sign convention, we get

PB = - u,

PB' = -v

PF = -f

PC = -2f

Now, In △ABC and △A'B'C, ∠ABC =∠A'B'C = 90°

∠ACB = ∠A'CB' (Vertically Opposite angles)

∴ △ABC ~△A'B'C (AA similarity)

$\frac{\mathrm{AB}}{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\frac{\mathrm{BC}}{\mathrm{B}\text{'}\mathrm{C}\text{'}}$ (the corresponding side of similar triangles are in proportion)..... (1)

In △DGF and △A'B'F, ∠DGF = ∠A'B'F = 90°

∠DGF= ∠A'FB' (vertically opposite angles)

△DGF ~△A'B'F (AA similarity)

$\frac{\mathrm{DG}}{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\frac{\mathrm{GF}}{\mathrm{B}\text{'}\mathrm{F}\text{'}}$(corresponding sides of similar triangles are in proportion)

But AB = DG (the perpendicular distance between two parallel lines are equal)

$\therefore \frac{\mathrm{AB}}{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\frac{\mathrm{GF}}{\mathrm{B}\text{'}\mathrm{F}\text{'}}....\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{eq}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{BC}}{\mathrm{B}\text{'}\mathrm{C}}=\frac{\mathrm{GF}}{\mathrm{B}\text{'}\mathrm{F}}...\left(3\right)$

Let us assume the mirror is very small,

∴ G and P are very close to each other so that GF = PF.

From equation (3),

$\frac{\mathrm{BC}}{\mathrm{B}\text{'}\mathrm{C}}=\frac{\mathrm{PF}}{\mathrm{B}\text{'}\mathrm{F}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{PB}-\mathrm{PC}}{\mathrm{PC}-\mathrm{PB}\text{'}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{PF}}{\mathrm{PB}\text{'}-\mathrm{PF}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-\mathrm{u}-(-2\mathrm{f})}{-2\mathrm{f}-(-\mathrm{v})}=\frac{{\displaystyle -\mathrm{f}}}{-\mathrm{v}-(-\mathrm{f})}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-\mathrm{u}+2\mathrm{f}}{-2\mathrm{f}+\mathrm{v}}=\frac{-\mathrm{f}}{-\mathrm{v}+\mathrm{f}}\phantom{\rule{0ex}{0ex}}\Rightarrow (-\mathrm{v}+\mathrm{f})(-\mathrm{u}+2\mathrm{f})\phantom{\rule{0ex}{0ex}}=-\mathrm{f}(-2\mathrm{f}+\mathrm{v})\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{vu}-2\mathrm{fv}-\mathrm{fu}+2{\mathrm{f}}^{2}\phantom{\rule{0ex}{0ex}}=2{\mathrm{f}}^{2}-\mathrm{fv}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{uv}=-\mathrm{fv}+2\mathrm{f}\phantom{\rule{0ex}{0ex}}\mathrm{v}+\mathrm{fu}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{uv}=\mathrm{vf}+\mathrm{uf}\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\mathrm{uvf}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{uv}}{\mathrm{uvf}}=\frac{\mathrm{vf}}{\mathrm{uvf}}+\frac{\mathrm{uf}}{\mathrm{uvf}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$

If the mirror is plane, the size of the image is always equal to the size of the object i.e., magnification is unity. But the case is different for a curved mirror. The size of the image is different from the size of the object in such a 'mirror'. The image may be greater or smaller in size than the object depending upon the nature of the mirror or the location of the object.

Let I and O be the size of the image and the object respectively. The ratio I/O is called magnification, and it is denoted by m.

Magnification, m = I/O = -v/u

This is called linear magnification.

Explain two advantages of a reflecting telescope over a refracting telescope.

- Spherical and chromatic abbreviation eliminated.
- Objective lenses are large and expensive in refracting telescope, where as reflecting telescope is economical.

44.

Define a wavefront. Using Huygens’ principle, verify the laws of reflection at a plane surface.

A wavefront is an imaginary surface over which an optical wave has a constant phase. The shape of a wavefront is usually determined by the geometry of the source.

Huygen's principle:

(i) Every point on a given wavefront acts as a fresh source of secondary wavelets which travel in all directions with the speed of light.

(ii) The forward envelope of these secondary wavelets gives the new wavefront at any instant.

Laws of reflection by Huygen's principle:

Let, PQ be reflecting surface and a plane wavefront AB is moving through the medium (air) towards the surface PQ to meet at the point B.

Let, c be the velocity of light and t be the time taken by the wave to reach A' from A.

Then, AA' = ct.

Using Huygen's principle, secondary wavelets starts from B and cover a distance ct in time t and reaches B'.

To obtain new wavefront, draw a circle with point B as centre and ct (AA' = BB') as radius. Draw a tangent A'B' from the point A'.

Then, A'B' represents the reflected wavelets travelling at right angle. Therefore, incident wavefront AB, reflected wavefront A'B' and normal lies in the same plane.

Consider ∆ABA' and B'BA'

AA' = BB' = ct [∵ AA' = BB' = BD = radii of same circle]

BA' = BA' [common]

∠BAA' = ∆BB'A' [each 90°]

∴ ∆ABA' ≅ ∠DBA' [by R.H.S]

∠ABA' = ∠B'A'B [corresponding parts of congruent triangles]

∴ incident angle i = reflected angle r

i.e., ∠i = ∠r

45.

In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ? Explain.

Size will be halved, intensity will be increased to four times.

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