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serious [3.7K]
3 years ago
8

A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, c

oncentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact?
A) a darkspotB) a bright spot thatis some color other than redC) a bright redspotD) a rainbow of color
Physics
2 answers:
Arte-miy333 [17]3 years ago
8 0

Answer:

The answer is: A) a darkspot

Explanation:

When the red light passes through the glass plate, it is scattered. the convex lens (convergent lens) has two refractive surfaces, therefore, in the center of the lens, a characteristic dark spot would be observed.

Korolek [52]3 years ago
6 0

Answer:

After passing through the glass plate, the red light disperses and meets at point.

The convex lens has two refracting surfaces, and convex kens is called as converging lens. So, at the exact center of the lens, one observes a Dark spot.

Thus, the correct option is a) one observes a dark spot.

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Read 2 more answers
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

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3 years ago
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