Light is incident on a prism with an equilateral triangular cross-section, making an angle of \theta w.r.t the normal as shown in the figure. The index of refraction of the prism is \displaystyle \frac{2}{\sqrt{3}} . If the angle \theta starts at 60^{\circ} and is gradually made smaller until no light exist from side BC, the final value of \theta in degrees is

A = 60^\circ r' + r = A So, \sin 90^\circ = \sqrt2 \sin r \Rightarrow r' = 45^\circ 45 + r = A \Rightarrow r = 15^\circ Since (1) \sin\theta = \sqrt2 \sin 15^\circ \sin\theta = \sqrt2 \sin 15^\circ \theta = \sin^{-1}(\sqrt2 \sin 15^\circ)

Ray Optics And Optical Instruments - Problem solving tips

Q: A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm.The distance from the lens a convex mirror of radius of curvature 10 cm should be placed on the other side of the lens such that the real images formed on the source itself is

for the final image to be formed at the source the image of the image through the lens should form at the central of the mirror. \dfrac{1}{\vartheta }-\dfrac{1}{\left ( -30 \right )} =\dfrac{1}{20} \dfrac{1}{\vartheta }=\dfrac{1}{20}-\dfrac{1}{30} \dfrac{1}{\vartheta }=\dfrac{10}{600} \vartheta =60cm d=60-10=50cm

Q: What should be the value of distance d so that final image is formed on the object itself. (focal lengths of the lenses are written on the lenses).

By lens maker formula \dfrac {1}{f}=\dfrac {1}{v}-\dfrac {1}{u}=(\dfrac {\mu_L}{\mu_M}-1)(\dfrac {1}{R_1}-\dfrac {1}{R_2}) note:- positive direction is direction of incident ray so all the values of u and v are taken accordingly with sign.For convex lens : \dfrac { 1 }{ f_1 } =\dfrac { 1 }{ v_1 } -\dfrac { 1 }{ u } \\ \dfrac { 1 }{ 10 } =\dfrac { 1 }{ v_1 } -\dfrac { 1 }{ (-10) } v_1=\infty For concave lens :Image of convex lens will now behave object for concave lens \dfrac { 1 }{ f_2 } =\dfrac { 1 }{ v_2} -\dfrac { 1 }{ v_1 } \\ \dfrac { 1 }{ -20 } =\dfrac { 1 }{ v_2 } -\dfrac { 1 }{ (\infty ) } \\ v_2=-20 cm 10+d=20 d=10

Q: A beam of light, consisting of red, green and blue colours, is incident on a right-angled prism, as shown. the refractive indices of the material of the prism for the above red, green and blue wavelengths are 1.39 1.44 and 1.47 respectively. The prism will :

The angle of incidence of all the rays is { 45 }^{ \circ } at the hypotenuse. For a critical angle of { 45 }^{ \circ } , the refractive index must be { \left( \sin { { 45 }^{ \circ } } \right) }^{ -1 }=\sqrt { 2 } =1.414 For red light, \mu =1.391.414. . Hence, their critical angles are <{ 45 }^{ \circ } . They will be reflected internally and emerge from the surface at the bottom.

Ray Optics And Optical Instruments

- Want to score better marks in exams? Have a look at some problem solving tips.

Tip 1
In problems involving a combination of lenses, always concentrate on one lens at a time.

For convenience, mark the crucial points such as f, 2f, R on the principal axis.
Start from the object, the direction of rays going towards the lens is taken to be positive.

We repeat the above procedure until no further images can be formed.

NOTE: The same tips are helpful while deriving the lens maker's formula as well as understanding the optical instruments like compound microscope and refracting telescopes.

Let us solve some sample questions:

A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm.The distance from the lens a convex mirror of radius of curvature 10 cm should be placed on the other side of the lens such that the real images formed on the source itself is