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kifflom [539]
3 years ago
14

The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which

it is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Physics
1 answer:
spayn [35]3 years ago
8 0

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

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Explanation:

It is given that,

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We know that the volume of the cylinder is given by :

V=\pi r^2 h

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h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
3 years ago
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katen-ka-za [31]
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N LC circuit has an oscillation frequency of 105 Hz. If C = 0.1 F , then L must be about:
Umnica [9.8K]

Answer:

L = 22.97 H

Explanation:

Given that,

Capacitance, C=0.1\ \mu F=0.1\times 10^{-6}\ F

Oscillation frequency, f = 0.5 Hz

The frequency of an AC circuit is given by :

f=\dfrac{1}{2\pi \sqrt{LC} }

Where

L is impedance

f^2=\dfrac{1}{4\pi ^2LC}\\\\L=\dfrac{1}{4\pi ^2 f^2 C}\\\\\text{Putting all the values}\\\\L=\dfrac{1}{4\pi^2 \times (105)^2\times 0.1\times 10^{-6}}\\\\L=22.97\ H

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Doppler's Effect: When a source moves with respect to the observer the frequency of the wave emitted from the source changes. If the source moves away from the observer, the frequency decreases and wavelength increases and vice versa.

Here the light source is moving towards the observer so the frequency will increase and wavelength will decrease. Thus the spectrum will shift towards the blue part. This is known as blue shift. The light wave will appear blue in color.

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3 years ago
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