To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

3 years ago

A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface

Blizzard [7]

Answer:

$\rho = 7.35\times \muC/m^3$

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

$E = \dfrac{Rl}{2\epsilon_0}$

ε₀ is the permitivity of free space

R is the radius of the rod

and also,

$\rho=\dfrac{2E\epsilon_0}{R}$

ρ is the volume charge density

$\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}$

$\rho = 7.35\times 10^{-6}\ C/m^3$

$\rho = 7.35\times \muC/m^3$

Hence, the volume charge density is equal to $\rho = 7.35\times \muC/m^3$

6 0

3 years ago