Light of wavelength 588.0 nm illuminates a slit of width 0.68 mm. (a) at what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.89 mm from the central maximum? m (b) calculate the width of the central maximum. mm

(a) 1.029 m

(b) 1.78 mm

Solution:

As per the question:

Wavelength of light,

Slit width, w = 0.68 mm

Now,

The distance first minimum in the diffraction pattern,

Now,

(a) For first minima:

where

D = Distance from the screen

D = 1.029 m

(b) The width of the central maxima is given by:

2y =

Explanation:

By definition, the density of the substance is given by:

D = \frac{m}{V}D=

V

m

Where,

m: mass of the substance

V: volume of the substance

Calculating the volume of the substance we have:

V = (50) * (10) * (30) V = 15000 cm ^ 3V=(50)∗(10)∗(30)V=15000cm

3

Then, replacing values we have:

D = \frac{3300}{15000}D=

15000

3300

D = 0.22 \frac{g}{cm^3}D=0.22

cm

3

g