\(Radius\ of\ curvature=40cm\\ focal\quad length=?\\ R=2f\\ 40=2\times f\\ f=\frac { 40 }{ 2 } =20cm\\ Focal\ length=20cm\)

The laws of refraction of light : The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a light of given colour and for the given pair of media. This law is also known as Snell's law of refraction.
 If i is the angle of incidence and r is the angle of refraction, then
\(\frac { \sin { i } }{ \sin { r } } =constant\)
Verification of laws of refraction :

(1) Fix a sheet of white paper on a drawing board using drawing pins.
(2) Place a rectangular glass slab over the sheet in the middle. Draw the outline of the slab with a pencil. Let us name the outline as ABCD. Take four identical pins.
(3) Fix two pins. Say E and F, vertically such that the line joining the pins is inclined to the edge AB.
(4) Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
(5) Remove the pins and the slab.
(6) Join the prints of the pins E and F and let it meet AB at O. Let EF meet AB at O. Similarly, join the prints of the pins G and H and Let it meet HG at O'. Join o and O'. Also produce EF as shown by a dotted line in Figure.
(7) Draw a perpendicular NN' to AB at O and another perpendicular MM' to CD at O'. In this activity, you will note that, the light ray has changed its direction at points O and O'. Note that both the points O and O' lie on surfaces separating two transparent media. The light ray has entered from air to glass and has bent towards the normal that is from a rarer to denser medium.
(8) The light ray has emerged from glass to air that is from a denser medium to a rarer medium. The light here has bent away from the normal. Compare the angle of incidence with the angle of refraction at both refracting surfaces AB and CD.
(9) In Figure, EO is the incident ray OO' the refracted ray and O'H the emergent ray. You may observe that the emergent ray is parallel to the direction of the incident ray.
(10) The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air-interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray.

(a) (i) At the centre of Curvature C

(ii) Between C and F

(iii) Between the focus F and the pole P of the mirror.

Speed of light in water \(=2.25 \times 10^{8} \mathrm{~ms}^{-1}\)
Speed of light in vacuum \(=3 \times 10^{8} \mathrm{~ms}^{-1}\)
(\(\mu\)) Refractive index of water =?
\(\mu =\frac{\text { Speed of light in vacuum }}{\text { Speed of light in water }}  \)
\(=\frac{3 \times \not10^{-8}}{2.25 \times \not10^{-8}}=\frac{3}{2.25} \)
\(\mu\) = 1.33

Distance of object = 7 cm
Magnified real image = distance of object x 3 times
= 7 x 3 = 21 cm
Location: In front of the mirror because it is concave mirror.