Ask question

Ask question

All categories

2 years ago

12

An optical engineer wishes to specify the properties of a concave makeup mirror, which is supposed to provide an upright and enl

arged image of the user's face. The engineer wants a magnification factor of two, and she assumes that the user's face will be 30 cm in front of the mirror. What radius of curvature should be specified (in cm) for the fabrication of these mirrors?

1 answer:

tangare [24]2 years ago

7 0

Answer:

radius of curvature is 1.2 m

Explanation:

given data

distance u = 30 cm

to find out

radius of curvature

solution

we know radius of curvature is double of focal length

so we find focal length

lens formula for focal length

1/f = 1/u + 1/v ..............................1

so here from magnification

M = - u /v

2 = - u /v )

v = -2u = -2(30)

v =- 60 cm

so from equation 1

1/f = 1/u + 1/v

1/f = 1/30 + 1/-60

f = 60.02 cm

so R = 2f

R = 2( 60.02)

R = 120.04 cm

so radius of curvature is 1.2 m

You might be interested in

A night-light uses 3 W.If the voltage is 120 V, what is the current? A.0.025A B.40 A C.360A D.0.25A

Tomtit [17]

The answer to this question is A.

6 0

2 years ago

A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me

enyata [817]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.

Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.

Explanation:

6 0

2 years ago

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

2 years ago

Where coastal rock is softer, waves erode the land faster

prohojiy [21]

Answer:

Headlands and bays are created where there are bands of hard and soft rock which meet the coastline at right angles. Softer rock is eroded more quickly and erodes backwards to form bays (which may have beaches). The harder rocks are more resistant to erosion and jut out into the sea to form exposed headlands

5 0

2 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

2 years ago