question_answer

(i) In a Young's double slit experiment, the two slits are kept 2 mm apart and the screen is positioned 140 cm away from the plane of the slits. The slits are illuminated with light of wavelength 600 nm. Find the distance of the third bright fringe, from the central maximum, in the interference pattern obtained on the screen.

(ii) If the wavelength of the incident light were changed to 480 nm, then find out the shift in the position of third bright fringe from the central maximum.

Or

A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.

Answer:

(i) Given, \[d=2\,mm,\,\,D=140\,\,cm,\,\,\lambda =600\,\,nm\]For bright fringes, \[x=n\,\frac{\lambda D}{d}\] For third bright fringe,                         \[x=\frac{3\times 600\times {{10}^{-9}}\times 140\times {{10}^{-2}}}{2\times {{10}^{-\,3}}}\] \[=126\times 10-5m=1.26mm\] (ii) When \[\lambda =480\,\,nm,\] then \[{{x}_{3}}=\frac{3\times 480\times {{10}^{-\,9}}\times 140\times {{10}^{-2}}}{2\times {{10}^{-3}}}\] \[=1008\times {{10}^{-\,6}}\] \[=100.8\times {{10}^{-5}}\,m\] Shift in bright fringe \[=126\times {{10}^{-\,5}}-100.8\times {{10}^{-\,5}}\] \[=25.2\times {{10}^{-5}}\,m\] \[=0.252\,\,mm\] Or For the convex lens \[f=+10\,\,cm,\,\,u=-\,30\,\,cm\] From the lens formula, \[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15}\] \[v=+15\,\,cm\] This image is at \[10\,\,cm\] from the concave lens which is placed at \[5\,\,cm\] from the convex lens. It will act as a virtual object. For concave lens \[u=+10\,\,cm,\,\,f=-10\,\,cm\] \[\therefore \] \[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{10}+\frac{1}{10}=0\] \[\Rightarrow \]   \[v=\infty \] Hence, the final image is formed at infinity. The image formation is shown below.