
In an experiment with a bi-prism, the readings on the optical bench of the position of the eyepiece and the two positions of the lens were respectively 1 m, 0.67 m and 34 cm. The distance between the two images for the two positions of the lens were respectively 0.3 mm and 1.2 mm and the width of 10 fringes was 9.72 mm. Assuming that there is no index error in any case, calculate:
A. The distance between the focal plane of interfering sources.
B. The wavelength of the light used.
Answer
571.5k+ views
Hint:
Optical bench is a system of devices which is used to determine the position of the image on the screen placed as per the setup and from this distance to calculate the focal length of the lens. Bi-prism is an optical device used to produce parallel rays of the same wavelength and intensity in order to fall on the screen.
Formula used:
$2d = \sqrt{d_1 \times d_2},\ \lambda = \dfrac{2d}{D}\times width$
Complete answer:
Given the distance between the eyepiece and second position of the lens is:
$V_1 =100 - 67 = 33 cm$
Hence, the reading of the slit on the bench is: $34 - 33= 1cm$
Hence the distance between the focal plane of the eyepiece and the plane of interfering sources is:
$D = 100 -1 = 99 cm = 0.99m$, which is the answer for A.
Now, we know the separation of two intersecting sources is:
$2d = \sqrt{d_1 \times d_2}$
Here, $d_1 = 1.2 mm = 0.12cm$
And $d_2 = 0.3mm = 0.03 cm$
Hence, $2d =\sqrt{0.12 \times 0.03} = 0.06cm$
Or $d=0.03cm$
Now, width of 10 fringes was 9.720 mm
Thus, 10 width = 9.72 mm
Or width = 0.972 mm.
Also, wavelength of light is given by $\lambda = \dfrac{2d}{D}\times width$
Hence putting the values:
$\lambda = \dfrac{2\times 0.03}{9.9} \times 0.0972 = 5891 \times 10^{-8} cm= 589.1 nm$
Which is the answer for part B.
Additional information:
The bi-prism is also called Fresnel’s bi-prism. It is mainly used to determine the interference fringes and to determine the wavelength of monochromatic light sources. But in optics, it has its importance mainly due to the fact that it is used to obtain coherent sources of light.
Note:
These types of experimental questions are frequently asked in many exams. Students need not to cram the formulas or procedure but try to understand the procedure and theory behind the experiment. Students are advised to practice all the experiments on their own and as many times as possible.
Optical bench is a system of devices which is used to determine the position of the image on the screen placed as per the setup and from this distance to calculate the focal length of the lens. Bi-prism is an optical device used to produce parallel rays of the same wavelength and intensity in order to fall on the screen.
Formula used:
$2d = \sqrt{d_1 \times d_2},\ \lambda = \dfrac{2d}{D}\times width$
Complete answer:
Given the distance between the eyepiece and second position of the lens is:
$V_1 =100 - 67 = 33 cm$
Hence, the reading of the slit on the bench is: $34 - 33= 1cm$
Hence the distance between the focal plane of the eyepiece and the plane of interfering sources is:
$D = 100 -1 = 99 cm = 0.99m$, which is the answer for A.
Now, we know the separation of two intersecting sources is:
$2d = \sqrt{d_1 \times d_2}$
Here, $d_1 = 1.2 mm = 0.12cm$
And $d_2 = 0.3mm = 0.03 cm$
Hence, $2d =\sqrt{0.12 \times 0.03} = 0.06cm$
Or $d=0.03cm$
Now, width of 10 fringes was 9.720 mm
Thus, 10 width = 9.72 mm
Or width = 0.972 mm.
Also, wavelength of light is given by $\lambda = \dfrac{2d}{D}\times width$
Hence putting the values:
$\lambda = \dfrac{2\times 0.03}{9.9} \times 0.0972 = 5891 \times 10^{-8} cm= 589.1 nm$
Which is the answer for part B.
Additional information:
The bi-prism is also called Fresnel’s bi-prism. It is mainly used to determine the interference fringes and to determine the wavelength of monochromatic light sources. But in optics, it has its importance mainly due to the fact that it is used to obtain coherent sources of light.
Note:
These types of experimental questions are frequently asked in many exams. Students need not to cram the formulas or procedure but try to understand the procedure and theory behind the experiment. Students are advised to practice all the experiments on their own and as many times as possible.
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