An object is placed at a distance 30cm from a thin lenses of power 4 diaptor. Discuss the nature of image

Physics

1 answer:

Alexxx [7]3 years ago

6 0

Answer:

nature of image = virtual

You might be interested in

if human beings ever travel to a planet whose mass and radius are both twice that of the earth, their weight on that planet will

irina [24]

The answer would be half their weight on Earth.

5 0

3 years ago

A ceiling fan draws a current of 0.625 A and has a voltage of 120 V. The resistance of the ceiling fan, to the nearest whole num

Fiesta28 [93]

So we want to know what is the resistance of the ceiling fan if the current I=0.625 A and the voltage V=120 V. So we can get the resistance from Ohms law: I=V/R where I is the current, V is voltage and R is resistance. So the resistance is R=V/I. Now we input the numbers and get: R=120/0.625=192 Ω. That is resistance to the closest whole number.

5 0

3 years ago

Read 2 more answers

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

4 years ago

A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source,

yaroslaw [1]

Answer:

1.58 W

Explanation:

Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is

$A = \pi r^2 = \pi 12^2 = 452.389 m^2$