# NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Q. Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict
the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [fig. (c)].

$$\textbf{Ans.}\space\text{From question for the glass-air interface :}\\\text{Angle of incidence,}i=60\degree\\\text{Angle of refraction,}r=35\degree\\\text{The relative refractive index of glass with respect to air is given by}\\_a\mu_g=\frac{\text{sin\space i}}{\text{sin}\space r}=\frac{\text{sin 60\degree}}{\text{sin}35\degree}\\ =\frac{0.8660}{0.5736}=1.51\space\text{…(i)}\\\text{For the air-water interface :}\\\text{Angle of incidence,}\text{i}=60\degree\\\text{Angle of refraction},r=47\degree\\\text{The relative refractive index of water with respect to air is given by Snell’s law as:}\\_w\mu_g=\frac{\text{sin}_i}{\text{sin}_r}=\frac{\text{sin}60\degree}{\text{sin}47\degree}\\=\frac{0.8660}{0.7314}=1.184\space\text{…(ii)}$$

From equations (i) and (ii), the relative refractive index of glass with respect to water can be obtained as:

$$_a\mu_g=\frac{_a\mu_g}{_a\mu_w}=\frac{1.51}{1.184}=1.275\\\text{The figure shows the situation involving the glass-water interface.}\\\text{Angle of incidence, i} = 45\degree\\ \text{Angle of refraction = r}\\\text{From Snell’s law, r can be calculated as :}\\\frac{\text{sin\space i}}{\text{sin\space r}}=_a\mu_g\\\frac{\text{sin\space 45\degree}}{\text{sin\space r}}=1.275\\\text{sin}\space r=\frac{\frac{1}{\sqrt{2}}}{1.275}=0.5546\\\therefore\space\text{r}=\text{sin}^{-1}(0.5546)=38.68\degree\\\text{Thus, the angle of refraction at the water-glass interface is 38.68}\degree$$

• Q. (i) If f = 0.5 m for a glass lens, what is the power of the lens ?
• (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass ?
• (iii) A convex lens has 20 cm focal length in air. What is focal length in water ? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5)
• Ans. (i) Power = + 2 dioptre.
• (ii) Given, f = + 12 cm, R1 = + 10 cm, R2 = – 15 cm.
• Refractive index of air is taken as unity.
• From lens formula as per sign convention has to be applied for f, R1 and R2.
• Substituting the values will give

$$\frac{1}{12}=(n-1)\bigg(\frac{1}{10}-\frac{1}{-15}\bigg)\\\text{This gives n = 1.5.}\\\text{(iii) For a glass lens in air,n}_2=1.5,n_1=1, f=+20\text{cm}.\text{Thus, the lens formula gives}\\\frac{1}{20}=(1.5-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)\\\text{For the same glass lens in water, n}_2 = 1.5, n_1 = 1.33. \text{Therefore,}\\\frac{1.33}{f}=(1.5-1.33)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)\\\text{Combining these two equations, we find f = + 78.2 cm.}$$

Q. Find the position of the image formed by the lens combination given in the fig.

Ans Image formed by the first lens (given by lens formula)

$$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\\\frac{1}{v_1}-\frac{1}{-30}=\frac{1}{10}\\\text{or}\space v_1=15\text{cm}$$

The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) cm = 10 cm to the right of the second lens.Though the image is real, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens.

$$\\\frac{1}{v_2}-\frac{1}{10}=\frac{1}{-10}\\\text{or}\space v_2=\infty\\\text{The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.}\\\frac{1}{v_3}-\frac{1}{u_3}=\frac{1}{f_3}\\\text{or}\space\frac{1}{v_3}-\frac{1}{u_3}=\frac{1}{f_3} \text{or}\space\text{v}_3=30\space \text{cm}\\\text{The final image is formed 30 cm to the right of the third lens.}$$

Q. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

• Ans. Given, focal length of the objective lens (fo) = 8 mm = 0.8 cm
• Focal length of the eyepiece (fe) = 2.5 cm
• Object distance for the objective lens (uo) = – 9.0 mm = – 0.9 cm
• Least distance of distant vision (d) = 25 cm
• Image distance for the eyepiece (ve) = – d = – 25 cm
• Let object distance for the eyepiece = ue
• From the lens formula, the value of ue as :

$$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\\\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\\=\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}=\frac{-11}{25}\\\therefore\space u_e=-\frac{25}{11}=-2.27\space\text{cm}\\\text{The value of the image distance for the objective lens (v}_o) \text{using the lens formula,}\\\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}\\\frac{1}{v_0}=\frac{1}{f_0}+\frac{1}{u_0}\\=\frac{1}{0.8}-\frac{1}{0.9}=\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}\\\therefore\space v_0=7.2\text{cm}\\\text{The distance between the objective lens and the eyepiece}=|u_e|+v_0\\=|2.27|+7.2=9.47\space\text{cm}\\\text{The magnifying power of the microscope is calculate as :}\\=\frac{v_0}{|u_0|}\bigg(1+\frac{d}{f_e}\bigg)\\=\frac{7.2}{0.9}\bigg(1+\frac{25}{2.5}\bigg)=8(1+10)\\=88\\\text{Thus, the magnifying power of the microscope is 88.}$$

Q. What should be the position of an object relative to biconvex lens so that it behaves like a magnifying lens ?

• Ans For biconvex lens to behave as a magnifying lens, the object must be placed between the focal point and optical centre of lens.