Image formed by converging lens. What is (a) the position, and (b) the size, of

Question

Image formed by converging lens. What is (a) the position, and (b) the size, of the image of a large 7.6-cm-high flower placed 1.00m from a +50.0-mm-focal-length camera lens? Diverging lens. Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm in front of the lens? Measuring the Wavelength of a Light Source A viewing screen is separated from a double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center [a) Determine the wavelength of the light.

Explanation / Answer

Answere 34.1)

As we know that

Camera lens is just a simple convex lens

so,

according to lens equation

1/o + 1/i = 1/f

So by putting values

we have

1/1 + 1/i = 1/0.5

1/i = 2-1 = 1

i = 1 m

magnification = i/o

magnification = 1/1 = 1

so

image the real image is 1*7.6 cm = 7.6 cm tall and inverted

The image is 7.6cm tall, is inverted and 1 m from the lens on the other side of the lens

Answer 34.3)

According to lens equation

1/i + 1/o = 1/f

o = object distance
i = image distance (negativce because virtual image)
f = focal length( negative because diverging lens)
1/i + 1/o = 1/f
1/-20 + 1/o = 1/-25
o = -100 cm

magnification equation:
M=-i/o
= - 20/100
= -0.2

Answer 37.1)

As we know that

y = m*lambda*D/d

Where

y = distance of fringe from center

m = order of fringe

D = distance between screen and slit

d = seperations between slits

So,

lambda = y*d / m*D

lambda = 4.5*10^-2*0.030*10^-3 / 2*1.2

lambda = 0.135*10^-5 / 2.4

lambda = 562.5*10^-9 m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---