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3 years ago

Would a liquid’s refractive index measured at a high altitude be different from that measured at sea level?

1 answer:

6 0

By definition, refractive index is the ratio of the velocity of light in vacuum to the velocity of light in a specified medium. The medium here is the liquid. The speed of light through the liquid also depends on the temperature and pressure. At sea level, the liquid is higher in temperature and pressure. So, the speed is gonna be relatively slower because of the movement of atoms. At high altitude, the liquid is at a lower temperature and pressure. The movements of the molecules are lesser, thus, light can move much faster. So, yes, it would be different. The refractive index would be closer to 1 at high altitude, and greater than 1 at sea level.

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Answer:

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3 years ago

A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. An

IRINA_888 [86]

Answer:

-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

0.023 m right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find image location i.e v2

so by lens formula v1 is

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

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

so from equation 1

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.35)

v2 = -0.233 m

so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

for Separation s = 45 cm = 0.45 m

v1 = 1/(1/f1 - 1/u1)

v1 =0.47 m

and

u2 = s - v1

u2 = 0.45 - 0.47 =- 0.02 m

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.02)

v2 = 0.023

so here 0.023 m right of diverging lens

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The three vectors in Fig. 3-33 have magnitudes a = 3.00 m, b = 4.00 m, and c = 10.0 m and angle θ = 30.0°. What are

aliya0001 [1]

Answer:

a) $a_{x} =3$ b) $a_{y} =0$

c) $b_{x} =3.46$ d) $b_{y} =2$

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g) p = -5.77 h) q=5

Explanation:

Diagram for given question is attached below in fig 1

for vector $\vec{a}$

θ = 0°

$a_{x} = 3 cos (0)\\a_{x} = 3\\a_{y} = 3 sin (0)\\a_{y} = 0$

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θ = 30°

$b_{x} = 4 cos (30)\\b_{x} = 3.46\\b_{y} = 4 sin (30)\\b_{y} = 2$

for vector $\vec{c}$

θ = 90°

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$\vec{c} = p\vec{a} + q\vec{b}$

$c =c_{x} \hat{i} + c_{y}\hat{j}\\as a_{y} =0\\c_{x} \hat{i} + c_{y}\hat{j} = pa_{x} \hat{i} +q(b_{x} \hat{i} +b_{y} \hat{j} )\\c_{x} \hat{i} = pa_{x} \hat{i} +qb_{x} \hat{i}\\c_{y}\hat{j} =qb_{y}\hat{j}$

$q=\frac{c_{y}}{b_{y}} \\q=\frac{10}{2}\\q=5$

$c_{x} \hat{i} = pa_{x} \hat{i} +qb_{x} \hat{i}\\0 = p(3) + (5)(3.46)\\p = -5.77$

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3 years ago