A light ray strikes a flat, L = 1.6 cm thick block (depth) of glass of refractive index 1.49 at an angle of 30 with the normal.

Calculate the lateral shift of the light ray, d.

there is a digram showing the light ray continuing at the same angle (30* to normal) until it exits the glass...so it has travelled down 1.6 cm.. once it leaves the angle decreases

is it correct to use

n1*sin(theta1)=n2*sin(theta2)

once i solve for sin(theta2)=opp/adj
sin(theta2)=d/1.6cm

Yes, it is correct to use the equation n1*sin(theta1) = n2*sin(theta2) for solving this problem. This equation, known as Snell's law, relates the angles of incidence and refraction of a light ray passing through an interface between two different media with different refractive indices.

In this case, the incident medium is air (or vacuum) with a refractive index of approximately 1 (n1 = 1), and the refractive medium is glass with a refractive index of 1.49 (n2 = 1.49). The incident angle is given as 30 degrees with respect to the normal.

To calculate the lateral shift (d), we need to find the outgoing angle of refraction (theta2). Rearranging Snell's law equation, we can solve for sin(theta2):

sin(theta2) = (n1*sin(theta1)) / n2

Plugging in the given values, we have:

sin(theta2) = (1*sin(30)) / 1.49
sin(theta2) = 0.642 / 1.49

Now, to find the value of sin(theta2), we can use the inverse sine function (sin^-1) on both sides of the equation:

theta2 = sin^-1(0.642 / 1.49)

After finding the value of theta2, we can use basic trigonometry to calculate the lateral shift (d). Since the light ray travels straight through the glass block, the lateral shift is equal to the horizontal distance traveled:

d = depth of the glass block * tan(theta2)

In this case, the depth of the glass block (L) is given as 1.6 cm. Plugging in the values and solving the equation will give you the lateral shift of the light ray.