Physics

7.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a * b.

0 Follower 25 Views

Contributor-Level 10

Let AB is equal to the vector a and AC be equal to the vector b.

Consider two vectors $\sin ? θ$ = $\frac{C N}{A C}$ = $\frac{C N}{|\overset{?}{B}|}$

$\overset{?}{A C}$ = $\overset{?}{b}$ inclined at an angle $θ$

MN = $\overset{?}{b} \sin ? θ$

| $\overset{?}{a}$ $* \overset{?}{b}$ | = | $\overset{?}{a}$ | | $\overset{?}{b} | \sin ? θ = A B * A C$

The area of ΔABC, we can write the relation

Area of Δ ABC = $\frac{1}{2}$ AB $* A C$ = $\frac{1}{2} [\overset{?}{a} * \overset{?}{b}]$

0 0

New answer posted

8 months ago

7.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

0 Follower 4 Views

Contributor-Level 10

The child is sitting on the trolley and there is no external force, hence it is a single system. The velocity of the centre of mass will not change, irrespective of any internal motion.

0 0

New answer posted

8 months ago

7.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

0 Follower 14 Views

Contributor-Level 10

Let us assume that H atom and Cl atom are at a distance of x and y respectively from the CM (Centre of Mass).

If mass of the H atom = m, mass of the Cl atom = 35.5m

Given x + y = 1,27 À

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore,

We can have, : (my+35.5mx)/ (m+35.5m) = 0

mx + 35.5my = 0

x = 35.5 (1.27 – x)

x = 1.24 À

So the centre of mass lies 1.24 À from H atom

0 0

New answer posted

8 months ago

7.1 Give the location of the centre of mass of a (i) Sphere (ii) Cylinder (iii) Ring and (iv) Cube each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

0 Follower 9 Views

Contributor-Level 10

All the structures specified are symmetric bodies with uniform mass density. For all these bodies, their centre of mass will lie in their geometric centres.

Not necessarily, the centre of gravity of a circular ring is at the imaginary centre of the ring.

0 0

New answer posted

8 months ago

9.32 Figure 9.34 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

0 Follower 81 Views

Contributor-Level 10

9.32 Focal length of the convex lens, $f_{1}$ = 30 cm

The liquid acts as a mirror, focal length of the liquid = $f_{2}$

Focal length of the system (convex lens + liquid), $f$ = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as

$\frac{1}{f}$ = $\frac{1}{f_{1}}$ + $\frac{1}{f_{2}}$ or $\frac{1}{f_{2}} = \frac{1}{f} - \frac{1}{f_{1}}$ = $\frac{1}{45}$ - $\frac{1}{30}$

$f_{2} =$ - 90 cm

Let the refractive index of the lens be $μ_{1}$ and the radius of curvature of one surface be R

Hence, the radius of curvature of the other surface is –R

R can be obtained by using the relation

$\frac{1}{f_{1}}$ = ( $μ_{1} - 1) (\frac{1}{R}$ + $\frac{1}{|R|}$ ) = (1.5 – 1)( $\frac{2}{R})$

$\frac{1}{30}$ = $\frac{1}{R}$ , so R =

...more

9.32 Focal length of the convex lens, $f_{1}$ = 30 cm

The liquid acts as a mirror, focal length of the liquid = $f_{2}$

Focal length of the system (convex lens + liquid), $f$ = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as

$\frac{1}{f}$ = $\frac{1}{f_{1}}$ + $\frac{1}{f_{2}}$ or $\frac{1}{f_{2}} = \frac{1}{f} - \frac{1}{f_{1}}$ = $\frac{1}{45}$ - $\frac{1}{30}$

$f_{2} =$ - 90 cm

Let the refractive index of the lens be $μ_{1}$ and the radius of curvature of one surface be R

Hence, the radius of curvature of the other surface is –R

R can be obtained by using the relation

$\frac{1}{f_{1}}$ = ( $μ_{1} - 1) (\frac{1}{R}$ + $\frac{1}{|R|}$ ) = (1.5 – 1)( $\frac{2}{R})$

$\frac{1}{30}$ = $\frac{1}{R}$ , so R = 30 cm

Let the refractive index of the liquid be $μ_{2}$

The radius of curvature of the liquid on the side of the plane mirror = $\infty$

Radius of curvature of the liquid on the side of the lens, R = -30 cm

The value of $μ_{2}$ can be calculated using the relation,

$\frac{1}{f_{2}}$ = ( $μ_{2} - 1) (\frac{1}{- R}$ - $\frac{1}{\infty}$ )

$\frac{- 1}{90}$ = ( $μ_{2} - 1) (\frac{1}{30}$ - 0)

$μ_{2}$ - 1 = $\frac{1}{3}$

$μ_{2} = \frac{4}{3}$ = 1.33

Hence, the refractive index of the liquid is 1.33.

less

0 0

New answer posted

8 months ago

9.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.33. A current in the coil produces a deflection of 3.5^o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

0 Follower 34 Views

Contributor-Level 10

9.31 Angle of deflection, $θ$ = 3.5 $°$

Distance of the screen from the mirror, D = 1.5 m

The reflected rays get deflected by an amount twice the angle of deflection, i.e. 2 $θ = 7 °$

The displacement (d) of the reflected spot of light on the screen is given as:

$\tan ? 2 θ = \frac{d}{D}$ = $\frac{d}{1.5}$

d = 1.5 tan 7 $°$ = 0.184 m = 18.4 cm

Hence, the deflection of the reflected spot of light is 18.4 cm.

0 0

New answer posted

8 months ago

9.30 A Cassegrain telescope uses two mirrors as shown in Fig. 9.30. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

0 Follower 6 Views

Contributor-Level 10

9.30 Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of objective mirror, $R_{1}$ = 220 mm

Hence focal length of the objective mirror, $f_{1}$ = $\frac{R_{1}}{2}$ = 110 mm

Radius of curvature of secondary mirror, $R_{2}$ = 140 mm

Hence focal length of the objective mirror, $f_{2}$ = $\frac{R_{2}}{2}$ = 70 mm

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror. Hence, the virtual object distance for the secondary mirror,

u = $f_{1} - d$ = 110 – 20 = 90 mm

Applying the mirror formula for the secondary mirror, we can cal

...more

9.30 Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of objective mirror, $R_{1}$ = 220 mm

Hence focal length of the objective mirror, $f_{1}$ = $\frac{R_{1}}{2}$ = 110 mm

Radius of curvature of secondary mirror, $R_{2}$ = 140 mm

Hence focal length of the objective mirror, $f_{2}$ = $\frac{R_{2}}{2}$ = 70 mm

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror. Hence, the virtual object distance for the secondary mirror,

u = $f_{1} - d$ = 110 – 20 = 90 mm

Applying the mirror formula for the secondary mirror, we can calculate the image distance ( $v$ ) as:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f_{2}}$ or $\frac{1}{v}$ = $\frac{1}{f_{2}} - \frac{1}{u}$ = $\frac{1}{70}$ - $\frac{1}{90}$

$v$ = 315 mm

Hence, the final image will be formed 315 mm away from the secondary mirror.

less

0 0

New answer posted

8 months ago

9.29 (a) For the telescope described in Exercise 9.28 (a), what is the separation between the objective lens and the eyepiece?

(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

(c) What is the height of the final image of the tower if it is formed at 25 cm?

0 Follower 46 Views

Contributor-Level 10

9.29 Focal length of the objective lens, $f_{o}$ = 140 cm

Focal length of the eyepiece, $f_{e}$ = 5 cm

Least distance of distinct vision, d = 25 cm

In normal adjustment, the separation between the objective lens and the eyepiece

= $f_{o} + f_{e} = 140 + 5 = 145 c m$

Height of the tower $h_{1} = 100 m$

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as : $θ = \frac{h}{u}$ = $\frac{100}{3000}$ = $\frac{1}{30}$ rad

The angle subtended by the image produced by the objective lens is given as $θ = \frac{h_{2}}{f_{o}}$ , where $h_{2}$ = height of the image of the tower formed by the objective lens

So, $h_{2}$ = $θ * f_{o}$ = $\frac{1}{30} * 140$ &nbs

...more

9.29 Focal length of the objective lens, $f_{o}$ = 140 cm

Focal length of the eyepiece, $f_{e}$ = 5 cm

Least distance of distinct vision, d = 25 cm

In normal adjustment, the separation between the objective lens and the eyepiece

= $f_{o} + f_{e} = 140 + 5 = 145 c m$

Height of the tower $h_{1} = 100 m$

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as : $θ = \frac{h}{u}$ = $\frac{100}{3000}$ = $\frac{1}{30}$ rad

The angle subtended by the image produced by the objective lens is given as $θ = \frac{h_{2}}{f_{o}}$ , where $h_{2}$ = height of the image of the tower formed by the objective lens

So, $h_{2}$ = $θ * f_{o}$ = $\frac{1}{30} * 140$ =4.7 cm

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation

m = 1 + $\frac{d}{f_{e}}$ = 1 + $\frac{25}{5}$ = 6

Height of the final image = m $* h_{2}$ = 6 $* 4.7 = 28.2 c m$

less

0 0

New answer posted

8 months ago

9.28 A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

(a) The telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) The final image is formed at the least distance of distinct vision (25 cm)?

0 Follower 9 Views

Contributor-Level 10

9.28 Focal length of the objective lens, $f_{o}$ = 140 cm

Focal length of the eyepiece, $f_{e}$ = 5 cm

Least distance of distinct vision, d = 25 cm

When the telescope is in normal adjustment, its magnifying power is given as:

$m_{1}$ = $\frac{f_{o}}{f_{e}}$ = $\frac{140}{5}$ = 28

When the final image is formed at d, the magnifying power of the telescope is given as:

$m_{2}$ = $\frac{f_{o}}{f_{e}} [1 + \frac{f_{e}}{d}$ ] = 28 $* [1 + \frac{5}{25}$ ] = 33.6

0 0

New answer posted

8 months ago

9.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

0 Follower 115 Views