A prism of refractive index $$n_1$$ and another prism of refractive index $$n_2$$ are stuck together (as shown in the figure). $$n_1$$ and $$n_2$$ depend on $$\lambda$$, the wavelength of light, according to the relation $$n_1 = 1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2}$$ and $$n_2 = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$$
The wavelength for which rays incident at any angle on the interface $$BC$$ pass through without bending at that interface will be _________ nm.

We need to find the wavelength ($$\lambda$$) for which light rays incident on the interface $$BC$$ pass through without bending (no deviation).

1. Condition for No Bending at the Interface

For a light ray to pass through an interface between two optical mediums without shifting or bending at any angle of incidence, the refractive indices of both mediums must be perfectly equal:

$$n_1 = n_2$$

2. Equate the Refractive Index Expressions

From the problem details , the equations for $$n_1$$ and $$n_2$$ are given as:

$$1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$$

Rearrange the terms to group the constant values on one side and the $\lambda$ terms on the other:

$$\frac{10.8 \times 10^{-14}}{\lambda^2} - \frac{1.8 \times 10^{-14}}{\lambda^2} = 1.45 - 1.2$$

$$\frac{(10.8 - 1.8) \times 10^{-14}}{\lambda^2} = 0.25$$

$$\frac{9 \times 10^{-14}}{\lambda^2} = 0.25$$

3. Solve for Wavelength ($$\lambda$$)

Isolate $$\lambda^2$$ in the equation:

$$\lambda^2 = \frac{9 \times 10^{-14}}{0.25}$$

$$\lambda^2 = 36 \times 10^{-14}\text{ m}^2$$

Taking the square root of both sides gives the value in meters:

$$\lambda = \sqrt{36 \times 10^{-14}} = 6 \times 10^{-7}\text{ m}$$

Convert the calculated metric value into nanometers ($1\text{ nm} = 10^{-9}\text{ m}$):

$$\lambda = 600 \times 10^{-9}\text{ m} = 600\text{ nm}$$

Conclusion

The wavelength for which the rays pass through the interface without bending is 600 nm.

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