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saul85 [17]
3 years ago
9

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

ed image with a magnification of 1.90?
Physics
1 answer:
alexandr402 [8]3 years ago
6 0

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

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iVinArrow [24]

Centripetal force and centrifugal force


4 0
3 years ago
A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on
svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}

Put the value into the formula

\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}

\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}

\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}

\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}

\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1

x=0.32\ m

Hence, The position on the x axis is 0.32 m.

5 0
3 years ago
Using Mirror equation A, Calculate The Frequency Of The Long Stand And The Shortest Wave Length Of That An Object Is Placed Of A
Leya [2.2K]

The Image distance and Magnification of The Image ​will be 30 cm and 3.

<h3>What is focal length?</h3>

The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.

Given data;

Focal length,f=?

Image distance,v=?

Object distance,u= 10 cm

Magnification,m= 2.85

The focal length is half of the radius;

f=R/2

f=30 Cm/2

f= 15 Cm

The mirror equation is found as;

\rm \frac{1}{f} =\frac{1}{v} +\frac{1}{u} \\\\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} }  \\\\\ v= -30 \ cm

The magnification of the lens is found as;

\rm m=\frac{30}{10}\\\\ m=3

Hence, the image distance and magnification of The image ​will be 30 cm and 3.

To learn more about the focal length refer;

brainly.com/question/16188698

#SPJ1

6 0
2 years ago
I need help plz and thank you❤
anzhelika [568]
I think:
In motion- 40
Not moving- 20
6 0
3 years ago
Does anyone know this one? Thanks
Inessa [10]

Answer:

3.844\,*\,10^5

So a=3.844 and b=5

Explanation:

Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

384,400=3.844 \,*\,100,000= 3.844 \,*\,10^5

with a=3.844, and "5" as the exponent of ten (so b=5)

6 0
3 years ago
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