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>> $Complex Numericals based on Thin Lenses$

Complex Numericals based on Thin Lenses

Q: Two identical glass \$(\mu_{g} = 3/2)\$ equiconvex lenses of focal length \$f\$ are kept in contact. The space between the two lenses is filled with water \$(\mu_{w} = 4/3)\$. The focal length of the combination is

Water in between the lens can be considered to be a concave lens of refractive index of water as shown in the figure. This suggests that the radius of curvature of all the lenses is same.Consider the convex lens-\$\dfrac{1}{f}=\dfrac{(\mu_{g}-1)2}{R}=\dfrac{(\dfrac{3}{2}-1)2}{R}\$hence \$R=f\$For the concave lens,\$Power=\dfrac{1}{f_{1}}=(\mu_{water}-1) (\dfrac{-2}{R})=\dfrac{-2}{3f}\$Power of the lenses get added,\$\dfrac{1}{f_{combination}}=\dfrac{1}{f}+\dfrac{-2}{3f}+\dfrac{1}{f}\$Hence \$f_{combination}=\dfrac{3f}{4}\$

Q: Drive an expression for equivalent focal length of two thin lenses kept in contact.

Let the focal lengths of the two lenses be \$f_1\$ and \$f_2\$.\$u_1 = OC_1\$ and \$v_1 = I_1C_1\$\$u_2= -C_1I_1\$ which is approximately equal to \$C_2 I_1\$ and \$v_2 = C_2I\$ which is approximately equal to \$C_1I\$ \$\therefore \frac{1}{u_1} + \frac{1}{v_1} = \frac{1}{f_1}\$ and \$ \frac{1}{u_2} + \frac{1}{v_2} = \frac{1}{f_2}\$ \$\frac{1}{OC_1} + \frac{1}{I_1C_1} = \frac{1}{f1}\$ and\$\frac {1}{- C_1I_1} + \frac{1}{C_1I} = \frac{1}{f_2}\$So \$\frac{1}{OC_1} + \frac{1}{I_1C_1} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{F}\$.Combined focal length of two thin lenses in contact is given by:\$F=\frac{f_1f_2}{f_1+f_2}\$

Q: A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of he combination, is :

Given,\$f_1=+40cm\$\$f_2=-25cm\$Power, \$P=\dfrac{1}{f}\$The power of combination,\$P=\dfrac{1}{f_1}+\dfrac{1}{f_2}\$\$P=\dfrac{100}{+40cm}+\dfrac{100}{-25cm}\$\$P=-1.5D\$The correct option is C.

Q: A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm apart from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.

Lens formula for convex lens,\$\cfrac{1}{v} - \cfrac{1}{u} = \cfrac{1}{f}\$\$\cfrac{1}{v} - \cfrac{1}{-60} = \cfrac{1}{20}\$\$v = 30\ cm\$Image (\$I_1\$) formed by the convex lens acts as an object for the convex mirror.Since the mirror and lens are separated by 15 cm, distance between \$I_1\$ and mirror is 15 cm.i.e. \$u = 15\$For convex mirror,\$f = \dfrac{R}{2} = 10\ cm\$From mirror equation,\$\cfrac{1}{v_2} + \cfrac{1}{u_2} = \cfrac{1}{f_2} \$\$\cfrac{1}{v_2} + \cfrac{1}{15} = \cfrac{1}{10}\$\$v_2 = 30\ cm\$Hence, image is formed 30 cm right of the convex mirror.From the ray diagram, it can be observed that the image is virtual.

Q: A plano convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to from the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object :

If the curved surface of a plano-convex lens is silvered, the focal length of the lens is given by\$\dfrac {1}{F} = \dfrac {2 \mu}{R}\$ \$\implies F = \dfrac {R}{2 \mu} = \dfrac {30}{2 \times 1.5} = 10 cm\$The real image will be equal to the size of the object if the object distance \$u = 2F = 2 \times 10 = 20 cm\$

Q: An object \$5 cm\$ is placed at a distance of \$20cm\$ in front of a convex mirror of radius of curvature \$30cm\$. Find the position of the image, its nature and size.

Formula:\$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\$Given,\$\dfrac{1}{v}+\dfrac{1}{(-20)}=\dfrac{1}{15}\$\$\dfrac{1}{v}=\dfrac{7}{60}\Rightarrow v=\dfrac{60}{7}\$\$v=8.57\approx 8.6cm\$The image is formed behind the mirror at a distance of \$8.6\ cm\$ \$m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}\$\$\rightarrow \dfrac{h_2}{5}=-\dfrac{8.57}{20}\$\$\therefore h_2=2.175\ cm\$Thus the image is \$2.175\ cm\$ high, virtual and erect.Hence, this is the answer.

Q: What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Focal length of convex lenses, \$f_1=30cm\$Focal length of concave lenses, \$f_2=-20cm\$\$\Rightarrow \dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}\$\$f=-60cm\$. Therefore diverging lens .

Q: Diameter of plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is \$2\times 10^{8}m/s\$, the focal length of the lens is:

The lens is as shown in the figure.To determine the radius of curvature \$ R \$ of the curved surface, we can write\$ R^2=3^2+(R-\cfrac{3}{10})^2 \$Solving this, we get\$ R=15.15 cm \$Speed of light in any medium is \$\dfrac{c}{\mu} \$ where \$ c=3\times 10^8 m/s \$ and \$ \mu \$ is the refractive index.For the lens, \$ \dfrac{c}{\mu}=2\times10^8 m/s \$\$ \mu=\dfrac{3\times10^8}{2\times10^8}=1.5 \$Putting these values in the lens makers formula for plano convex lens, we have,\$\dfrac{1}{f}=(\mu-1)(\dfrac{1}{R})=(1.5-1)(\dfrac{1}{15.15}) cm^{-1} \$Solving this, we get \$ f\approx 30 cm \$

Q: Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is \$F_1\$. When the space between the two lenses is filled with glycerine (which has the same refractive index (\$\mu = 1.5\$) as that of glass) then the equivalent focal length is \$F_2\$. The ratio \$F_1 : F_2\$ will be :

Equivalent focal length in air,\$\dfrac{1}{F_1} = \dfrac{1}{f} + \dfrac{1}{f} = \dfrac{2}{f}\$ When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (-f)\$\implies \dfrac{1}{F_2} = \dfrac{1}{f} + \dfrac{1}{f} - \dfrac{1}{f}\$\$ = \dfrac{1}{f}\$\$\implies \dfrac{F_1}{F_2} = \dfrac{1}{2}\$

Q: Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

\$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}\$\$\frac{1}{f_1}=(\mu-1)(\frac{1}{\infty}-\frac{1}{-R})=(1.5-1)(\frac{1}{\infty}+\frac{1}{R})\Rightarrow f_1=2R\$\$\frac{1}{f_2}=(\mu-1)(\frac{1}{-R}-\frac{1}{R})=-(1.7-1)(\frac{1}{R}+\frac{1}{R})\Rightarrow f_2=-\frac{5R}{7}\$\$\frac{1}{f_3}=(\mu-1)(\frac{1}{R}-\frac{1}{R})=(1.5-1)(\frac{1}{R}+\frac{1}{\infty})\Rightarrow f_3=2R\$\$\frac{1}{f_{eq}}=\frac{1}{2R}-\frac{7}{5R}+\frac{1}{2R}\$\$f_{eq}=-2.5R=-50 cm\$

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