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Fofino [41]
3 years ago
12

A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv

e and the eyepiece are separated by 19.6 cm. The final image is at infinity. a) What is the angular magnification? (b) How far from the objective should the object be placed.
Physics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

so

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

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Vadim26 [7]

Answer:

d = 5.75m

Explanation:

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n1= refractive index of 1st medium= 1

n2=  refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

r = \sin^{-1}\frac{n_1\sin i}{n_2}

Here,

i = 90 - θ

\theta = \tan^-^1(\frac{1.5}{3} )\\\\=26.56^\circ

r = \sin^{-1}\frac{n_1\sin i}{n_2}

r = \sin^{-1}\frac{(1)\sin (90-26.56)}{1.33}\\\\r = 42.26m

\tan r = \frac{2.5}{d_1}

d_1 = \frac{2.5}{\tan (42.26)} \\\\d_1 = 2.75m

Therefore, the distance is

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6 0
3 years ago
A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting
Alisiya [41]

The magnitude of the work done by the force of friction is 1045.68 joules

Explanation:

Work done = force × distance

  • Work done by friction force (W) = friction force (F) × distance (d)
  • Friction force (F) = coefficient of  kinetic friction (μ) × normal reaction force (R)

A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting  at an angle of 25.8. above the horizontal. The block is displaced 57,8 m, and the coefficient of  kinetic friction is 0.229

To find the normal reaction force of the block distribute The constant force into two component, horizontal component of 132 cos(25.8°) and vertical component 132 sin(25.8°), Then equate ∑vertical forces by 0 to find R

The weight of the block = mg, where m is its mass and g is 9.8 m/s²

→ ∑Vertical forces = R + 132 sin(25.8°) - mg

→ ∑Vertical forces = R + 132 sin(25.8°) - (16.7)(9.8)

→ ∑Vertical forces = R - 106.2

Equate it by 0

→ R - 106.2 = 0 ⇒ add 106.2 to both sides

→ R = 106.2 N

Now let us find the friction force

→ F = μ R

→ μ = 0.229

→ F = 0.229 (106.2) = 24.3198 N

Let us calculate the work don by the force of friction

→ W = F × d

→ d = 57.8 meters

→ W = 24.3198 × 57.8 = 1045.68 joules

The magnitude of the work done by the force of friction is 1045.68 joules

Learn more:

You can learn more about force of friction in brainly.com/question/6217246

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